Paper Questions (Raw)
Q1. The coordinates of a moving particle at time t are given by (x=ct^{3}) and (y=bt^{3}). The speed of the particle, (at t=1) s, is
1) 3 (\sqrt{c^{2}+b^{2}})
2) 3 ((c+b))
3) (\frac{1}{3} \sqrt{c^{2}+b^{2}})
4) 3 (\sqrt{c^{2}-b^{2}})
Hint: At
(t=1 s)$$v_{X}=\frac{d x}{d t}=3 c t^{2}=3 c \\v_{Y}=\frac{d y}{d t}=3 b t^{2}=3 b \\ \therefore v_{\text {net }}=\sqrt{v_{X}^{2}+v_{Y}^{2}}=\sqrt{(3 c)^{2}+(3 b)^{2}}=3 \sqrt{c^{2}+b^{2}} m / s$$
Q2. Five balls (A, B, C, D) and E are projected with the same speed making angles (10^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}) and (80^{\circ}) respectively with the horizontal. Which ball will strike the ground at the farthest point?
Hint: The maximum range is obtained for (\theta=45^{\circ})
Q3. On which of the following, is the maximum height achieved by a projectile, independent?
1) vertical component of the initial velocity
2) Angle of projection
3) Acceleration due to gravity
4) horizontal component of initial velocity
Q4. The angle of projection for same maximum range is
1) (45^{\circ})
2) (53.5^{\circ})
3) (63^{\circ})
4) (73^{\circ})
Q5. If (R) is the range of a projectile motion, then equation of its trajectory is
1) (y=x \cot \theta\left(1-\frac{x}{R}\right))
2) (y=x \cot \theta\left(1+\frac{x}{R}\right))
3) (y=x \tan \theta\left(1-\frac{x}{R}\right))
4) (y=x \tan \theta\left(1+\frac{x}{R}\right))
Q6. In a projectile motion, the velocity
1) is always perpendicular to the acceleration.
2) is never perpendicular to the acceleration.
3) is perpendicular to the acceleration for one instant only.
4) is perpendicular to the acceleration for two instants.
Q7. Two projectiles are fired with same speed with two different angles of projections, so as to have same range (R). If the maximum height reached by them are (h_{1}) and (h_{2}), then (R=)
1) (\frac{4 h_{1} h_{2}}{h_{1}+h_{2}})
2) (\frac{4 h_{1} h_{2}}{h_{1}-h_{2}})
3) (4 \sqrt{h_{1} h_{2}})
4) (2 \sqrt{h_{1} h_{2}})
Hint: Same range is obtained for an angles
(\theta) and
(90-\theta).
$$h_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g} \\h_{2}=\frac{u^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 g}=\frac{u^{2} \cos ^{2} \theta}{2 g} \\R=u^{2}\left(\frac{\sin 2 \theta}{g}\right)=\frac{2(u \sin \theta)(u \cos \theta)}{g}=4\left(\sqrt{\frac{u^{2} \sin ^{2} \theta}{2 g}}\right)\left(\sqrt{\frac{u^{2} \cos ^{2} \theta}{2 g}}\right)=4 \sqrt{h_{1} h_{2}}$$
Q8. A particle is projected with a velocity v so that its horizontal range is twice the greatest height attained. The horizontal range is
1) (\frac{4 v^{2}}{5 g})
2) (\frac{v^{2}}{g})
3) (\frac{v^{2}}{2 g})
4) (\frac{2 v^{2}}{3 g})
Hint: (R=2 H)$$\frac{v^{2} \sin 2 \theta}{g}=2 \times \frac{v^{2} \sin ^{2} \theta}{2 g}\\2 \sin \theta \cos \theta=\sin ^{2} \theta\\ \tan \theta=2\\ \therefore \sin \theta=\frac{2}{\sqrt{5}} \text { and } \cos \theta=\frac{1}{\sqrt{5}}\\R=\frac{v^{2} \times \sin 2 \theta}{g}=v^{2} \times \frac{2 \sin \theta \cos \theta}{g}=v^{2} \times \frac{2}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}\\ \therefore R=\frac{4 v^{2}}{5 g}$$
Q9. A ball is projected with a speed of (40 m / s) at an angle (45^{\circ}) with horizontal. There is a wall of 50 m height at a distance of 40 m from the projection point. The ball will hit the wall at a height of
1) 10 m
2) 20 m
3) 30 m
4) 40 m
Hint: $$y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}=40 \tan 45^{\circ}-\frac{10(40)^{2}}{2(40)^{2} \cos ^{2} 45^{\circ}}=40-10=30 m$$
Q10. A particle moves in the X-Y plane according to the law (x=kt) and (y=kt(1-\alpha t)), where k and (\alpha) are positive constants and (t) is time. What is the equation of trajectory of the particle?
1) (y=k x)
2) (y=x-\frac{\alpha x^{2}}{k})
3) (y=-\frac{\alpha x^{2}}{k})
4) (y=\alpha x)
Hint: (x=u \cos \theta t=kt)(u \cos \theta=k)(y=u \sin \theta t-\frac{1}{2} g t^{2}=kt-k \alpha t^{2})(u \sin \theta=k)(\frac{u \sin \theta}{u \cos \theta}=\tan \theta=1)(g=2 k \alpha)Equation of trajectory,
$$y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}=x-\frac{2 k \alpha x^{2}}{2 k^{2}}=x-\frac{\alpha x^{2}}{k}$$
Q11. For a projectile motion, if (x=8 t) and (y=2 t-3 t^{2}), then its time of flight is
1) (\frac{2}{3} s)
2) (\frac{1}{3} s)
3) (\frac{4}{3} s)
4) (\frac{5}{3} s)
Hint: For maximum height,(v_{y}=2-6 t=0)(t=\frac{1}{3} s=T_{a})(T=2 T_{a}=\frac{2}{3} s)
Q12. A shell fired from a canon can cover maximum horizontal distance of 10 km . Then velocity of projection is
1) (\sqrt{980} m / s)
2) (\sqrt{9800} m / s)
3) (\sqrt{98000} m / s)
4) (10^{2} \sqrt{98} m / s)
Hint: (R_{\max }=\frac{u^{2}}{g}=10000)(u^{2}=9.8 \times 10000)(u^{2}=98000)(u=\sqrt{98000} m / s)
Q13. A particle of mass m is projected with velocity v making an angle of (45^{\circ}) with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be
1) (\frac{mv}{\sqrt{2}})
2) (\sqrt{2} mv)
3) 0
4) 2 mv
Hint: The horizontal momentum does not change.The change in vertical momentum is(\Delta p=mv \sin 45-(-mv \sin 45)=\frac{2 mv}{\sqrt{2}}=\sqrt{2} mv)
Q14. Two boys stationed at A and B fire bullets simultaneousllly at a bird stationed at C . The bullets are fired from A and B at an angles of (53^{\circ}) and (37^{\circ}) respectively with the vertical. Both the bullets hit the bird simultaneously. If (v_{B}=60 m / s), then (v_{A}=(\tan 37=\frac{3}{4}))
1) (80 m / s)
2) (40 m / s)
3) (45 m / s)
4) (90 m / s)
Hint: To hit the bird simultaneously, the vertical component must be same.
(\therefore v_{A} \cos 53=v_{B} \cos 37)$$v_{A}=v_{B} \frac{\cos 37}{\cos 53}=v_{B} \frac{\cos 37}{\cos (90-53)}=v_{B} \frac{\cos 37}{\sin 37}=\frac{v_{B}}{\tan 37}=60 \times \frac{4}{3}=80 m / s$$
Q15. A stone is projected with a velocity (20 \sqrt{2} m / s) at an angle of (45^{\circ}) to the horizontal. The average velocity of stone during its motion from starting point to its maximum height is (take (g=10 m / s^{2}))
1) (20 m / s)
2) (20 \sqrt{5} m / s)
3) (5 \sqrt{5} m / s)
4) (10 \sqrt{5} m / s)
Hint: $$H=\frac{v^{2} \sin ^{2} \theta}{2 g}=\frac{(20 \sqrt{2})^{2} \sin ^{2} 45}{2 \times 10}=\frac{800 \times \frac{1}{2}}{2 \times 10}=20 m\\R=\frac{v^{2} \sin 2 \theta}{g}=\frac{(20 \sqrt{2})^{2} \sin (2 \times 45)}{10}=\frac{800 \times 1}{10}=80 m\\ \text {Displacement} =\sqrt{H^{2}+\left(\frac{R}{2}\right)^{2}}=\sqrt{(20)^{2}+(40)^{2}}=\sqrt{400+1600}=\sqrt{2000}=20 \sqrt{5} m\\v_{av}=\frac{\text {Displacement}}{\text {Time}}=\frac{20 \sqrt{5}}{2}=10 \sqrt{5} m / s$$
Q16. If 2 balls are projected at angles (45^{\circ}) and (60^{\circ}) and the maximum heights reached are same, what is the ratio of their initial velocities?
1) (\sqrt{2}: \sqrt{3})
2) (\sqrt{3}: \sqrt{2})
3) 3: 2
4) 2: 3
Hint: (H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g} \Rightarrow H \propto u^{2} \sin ^{2} \theta=\text{same})$$\frac{u_{1}}{u_{2}}=\frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{\sin 60}{\sin 45}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$$
Q17. The angle of projection, for which the horizontal range and the maximum height of a projectile are in the ratio of (2: 1) is
1) (45^{\circ})
2) (\theta=\tan ^{-1}(0.25))
3) (\theta=\tan ^{-1}(2))
4) (60^{\circ})
Hint: According to given condition,
(R=2 H)$$\frac{u^{2} \sin 2 \theta}{g}=\frac{2 u^{2} \sin ^{2} \theta}{2 g}\\2 \sin \theta \cos \theta=\frac{2 \sin ^{2} \theta}{2}\\ \tan \theta =2 \\ \theta =\tan ^{-1}(2)$$
Q18. The equation of a path of a projectile is (y=\frac{x}{\sqrt{3}}-\frac{1}{2} gx^{2}). The angle of projection with the vertical is
1) (30^{\circ})
2) (60^{\circ})
3) (45^{\circ})
4) (53^{\circ})
Hint: (y=(\tan \theta) x-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}=\frac{x}{\sqrt{3}}-\frac{1}{2} g x^{2})$$\tan \theta =\frac{1}{\sqrt{3}} \\ \theta =30^{\circ}$$
Q19. A body is projected with initial velocity of ((8 \hat{i}+6 \hat{j}) m / s). The horizontal range is
1) 9.6 m
2) 14 m
3) 50 m
4) 24.6 m
Hint: $$R=\frac{u^{2} \sin 2 \theta}{g}=\frac{2(u \sin \theta)(u \cos \theta)}{g}=\frac{2 \times 8 \times 6}{10}=9.6 m$$
Q20. The range of a particle when launched at an angle of (15^{\circ}) with the horizontal is 150 m . What is the range of projectile, when launched at an angle of (45^{\circ}) to the horizontal?
1) 150 m
2) 300 m
3) 450 m
4) 600 m
Hint: $$R_{1}=\frac{u^{2} \sin 2 \theta_{1}}{g} \\150=\frac{u^{2} \sin 30}{g} \\ \frac{u^{2}}{g}=300 \\R_{2}=\frac{u^{2} \sin 2 \theta_{2}}{g}=\frac{u^{2} \sin 90}{g}=300 m$$
Q21. Two stones having different masses (m_{1}) and (m_{2}) are projected at angles (\theta) and ((90^{\circ}-\theta)) with same velocity from the same point. The ratio of their maximum heights is
1) 1: 1
2) 1: (\tan \theta)
3) (\tan \theta: 1)
4) (\tan ^{2} \theta: 1)
Hint: Maximum height for mass
(m_{1}, H_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g})Maximum height for
(m_{2}, H_{2}=\frac{u^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 g})$$\therefore \frac{H_{1}}{H_{2}}=\frac{\sin ^{2} \theta}{\sin ^{2}\left(90^{\circ}-\theta\right)}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{\tan ^{2} \theta}{1} \\ \therefore H_{1}: H_{2}:: \tan ^{2} \theta: 1$$
Q22. The equation of path of a projectile is (y=x-20 x^{2}). The maximum height reached by the body, above the point of projection is
1) (\frac{1}{80})
2) (\frac{1}{1600})
3) (\frac{1}{40})
4) (\frac{1}{320})
Hint: (y=x-20 x^{2})For
(y=0),
(x-20 x^{2}=0)(x=R=\frac{1}{20})For
(y=H_{m}, x=\frac{R}{2}=\frac{1}{40})$$H_{m}=\left[\frac{1}{40}\right]-\left[20 \times\left(\frac{1}{40}\right)^{2}\right]=\frac{1}{40}-\frac{1}{80}=\frac{1}{80}$$
Q23. A stone is projected from the ground with velocity (50 m / s) at an angle of (30^{\circ}). It crosses a wall after 3 s . How far beyond the wall the stone will strike the ground? [ (g=10 m / s^{2}))
1) 90.2 m
2) 89.6 m
3) 86.6 m
4) 70.2 m
Hint: (T=\frac{2 u \sin \theta}{g}=\frac{2 \times 50 \times \frac{1}{2}}{10}=5 s)Time to cross the wall
(=3 s) (given)Time in air after crossing the wall
(=(5-3)=2 s)∴ Distance travelled beyond the wall
$$x=(u \cos \theta) t=50 \times \frac{\sqrt{3}}{2} \times 2=86.6 m$$
Q24. Two projectiles, one fired from the surface of the earth with speed (5 m / s) and other fired from the surface of a planet with initial speed of (3 m / s), traces identical trajectories. Neglecting friction effect, the value of acceleration due to gravity on the planet is
1) (5.9 m / s^{2})
2) (3.5 m / s^{2})
3) (16.3 m / s^{2})
4) (8.5 m / s^{2})
Hint: Here,
(\frac{u_{e}^{2}}{g_{e}}=\frac{u_{p}^{2}}{g_{p}} \quad(\text{Range}=\text{constant}))$$g_{p}=g_{e} \times\left(\frac{u_{p}}{u_{e}}\right)^{2}=9.8 \times\left(\frac{3}{5}\right)^{2}=\frac{9.8 \times 9}{25}=0.392 \times 9=3.528 m / s^{2}$$
Q25. From the top of a tower, three balls (A, B) and C whose mass is in the ratio (1: 2: 3) are thrown with the same speed (\mathrm{v}_{0}) as shown. The ratio of speed with which they strike the ground is
1) 1: 2: 3
2) 1: 1: 1
3) 3: 2: 1
4) 1: 4: 9
Hint: (K_{1}+U_{1}=K_{2}+U_{2})(\frac{1}{2} m v_{0}^{2}+m g h=\frac{1}{2} m v^{2})(v=\sqrt{v_{0}^{2}+2 g h})(As v_{0} \text{ and } h \text{ are same for all three cases, } v_{A}=v_{B}=v_{C})
Q26. A body, of mass 4 kg , is fired with velocity ((5 \hat{i}+4 \hat{j}) ms^{-1}). Its minimum kinetic energy will be
1) (50 \sqrt{2} J)
2) 100 J
3) 50 J
4) (25 \sqrt{2} J)
Hint: At highest position, we get minimum kinetic energy with velocity
(u \cos \theta=5 m / s.)$$K_{\min }=\frac{1}{2} m(u \cos \theta)^{2}=\frac{1}{2} \times 4 \times 25=50 J$$
Q27. The equation of a projectile path is given by, (y=\sqrt{3} x-x^{2}). The range of motion is
1) 3 m
2) (3 \sqrt{3} m)
3) (\sqrt{3} m)
4) (1 / \sqrt{3} m)
Hint: Standard equation of projectile motion
$$y=(\tan \theta) x-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}=\sqrt{3} x-x^{2}$$
$$\frac{2 u^{2} \cos ^{2} \theta}{g}=1$$
(\tan \theta=\sqrt{3})$$R=\frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2}(2 \sin \theta \cos \theta)}{g}=\left(\frac{2 u^{2}}{g}\right)(\sin \theta \cos \theta)=\frac{\sin \theta \cos \theta}{\cos ^{2} \theta}=\tan \theta=\sqrt{3} m$$
Q28. The equation of a projectile path is given by, (y=\sqrt{3} x-x^{2}). Maximum height reached by the body is
1) (\sqrt{3} m)
2) 0.75 m
3) 0.5 m
4) (\frac{\sqrt{3}}{2} m)
Hint: $$H=\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{u^{2}\left(3 \cos ^{2} \theta\right)}{2 g}=\frac{3}{4}=0.75 m$$
Q29. The equation of a projectile path is given by, (y=\sqrt{3} x-x^{2}). The angle of projection is
1) 75
2) (30^{\circ})
3) (60^{\circ})
4) (45^{\circ})
Hint: (\tan \theta = \sqrt{3} \Rightarrow \theta = 60^{\circ})
Q30. The equation of a projectile path is given by, (y=\sqrt{3} x-x^{2}). The velocity at the highest point is
1) (\sqrt{5} m / s)
2) (\sqrt{3} m / s)
3) (\sqrt{10} m / s)
4) (\frac{\sqrt{5}}{2} m / s)
Hint: $$v=u \cos \theta=\sqrt{\frac{g}{2}}=\sqrt{5} m / s$$
Q31. The equation of a projectile path is given by, (y=\sqrt{3} x-x^{2}). The velocity of projection is
1) (\frac{\sqrt{5}}{2} m / s)
2) (\sqrt{3} m / s)
3) (\sqrt{10} m / s)
4) (2 \sqrt{5} m / s)
Hint: $$u=\frac{\sqrt{5}}{\cos 60}=2 \sqrt{5} m / s$$
Q32. Three projectiles are fired at an angle of projection of (20^{\circ}, 45^{\circ}, 70^{\circ}), with same velocity. Their range is X , Y and Z . Then
1) (X=Y
2) (X=Z>Y)
3) (X=Y>Z)
4) (X=Z
Hint: (R=\frac{u^{2} \sin 2 \theta}{g} \Rightarrow R \propto \sin 2 \theta)(\therefore X=Z
Q33. Three projectiles are fired at an angle of projection of (20^{\circ}, 45^{\circ}, 70^{\circ}), with same velocity. If the maximum heights reached by the projectiles are (H_{1}, H_{2}) and (H_{3}), then (H_{1}+H_{3}=)
1) (H_{2} / 2)
2) (2 H_{2})
3) (4 H_{2})
4) (H_{2})
Hint: (H_{2}=\frac{u^{2} \sin ^{2} 45}{2 g}=\frac{u^{2}}{4 g})(H_{1}=\frac{u^{2} \sin ^{2} \theta_{1}}{2 g})(H_{3}=\frac{u^{2} \sin ^{2}\left(90^{\circ}-\theta_{1}\right)}{2 g}=\frac{u^{2} \cos ^{2} \theta_{1}}{2 g})$$H_{1}+H_{3}=\frac{u^{2} \sin ^{2} \theta_{1}}{2 g}+\frac{u^{2} \cos ^{2} \theta_{1}}{2 g}=\frac{u^{2}}{2 g}=2 H_{2}$$
Q34. Three projectiles are fired at an angle of projection of (20^{\circ}, 45^{\circ}, 70^{\circ}), with same velocity. If the time of flights of the projectiles are (T_{1}, T_{2}) and (T_{3}), then (T_{1}^{2}+T_{3}^{2}=)
1) (T_{2}^{2} / 2)
2) (2 T_{2}^{2})
3) (4 T_{2}^{2})
4) (T_{2}^{2})
Hint: (T_{2}=\frac{2 u \sin \theta_{2}}{g})(\Rightarrow T_{2}^{2}=\frac{4 u^{2} \sin ^{2} 45}{g}=\frac{2 u^{2}}{g})(T_{1}=\frac{2 u \sin \theta_{1}}{g})(\Rightarrow T_{1}^{2}=\frac{4 u^{2} \sin ^{2} \theta_{1}}{g})(T_{3}=\frac{2 u \sin \theta_{3}}{g})(\Rightarrow T_{3}^{2}=\frac{4 u^{2} \sin ^{2}\left(90-\theta_{1}\right)}{g}=\frac{4 u^{2} \cos ^{2} \theta_{1}}{g})$$T_{1}^{2}+T_{3}^{2}=\frac{4 u^{2} \sin ^{2} \theta_{1}}{g}+\frac{4 u^{2} \cos ^{2} \theta_{1}}{g}=\frac{4 u^{2}}{g}=2 T_{2}^{2}$$
Q35. A body is projected with a velocity of (10 m / s) at angle of projection of (60^{\circ}). Its velocity at the end of (1 / \sqrt{3} s) is ((g=10 m / s^{2}))
1) (5 \hat{i}+5 \sqrt{3} \hat{j} m / s)
2) (5 \hat{i}+\frac{5 \sqrt{3}}{3} \hat{j} m / s)
3) (5 \hat{i}+\frac{5 \sqrt{3}}{2} \hat{j} m / s)
4) (5 \hat{i}+\frac{5 \sqrt{3}}{4} \hat{j} m / s)
Hint: (T=\frac{2 u \sin \theta}{g}=\frac{2 \times 10 \times \frac{\sqrt{3}}{2}}{10}=\sqrt{3} s)$$v_{x}=u \cos 60=10 \times 0.5=5 m / s \\v_{y}=u \sin 60-g t=\left(10 \times \frac{\sqrt{3}}{2}\right)-\left(10 \times \frac{1}{\sqrt{3}}\right)=5 \sqrt{3}-\frac{10}{\sqrt{3}}=\frac{5}{\sqrt{3}}=\frac{5 \sqrt{3}}{3} m / s \\ \vec{v}=v_{x} \hat{i}+v_{y} \hat{j}=5 \hat{i}+\frac{5 \sqrt{3}}{3} \hat{j} m / s$$
Q36. A body is projected with a velocity of (10 m / s) at angle of projection of (60^{\circ}). The velocity of the body when it attains a height of (\frac{15}{4} m) is
1) (5 \hat{i}-5 \sqrt{3} \hat{j} m / s)
2) (5 \hat{i} m / s)
3) (5 \hat{i}+\frac{5 \sqrt{3}}{2} \hat{j} m / s)
4) (5 \hat{i}+\frac{5 \sqrt{3}}{4} \hat{j} m / s)
Hint: (H=\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{100 \times \frac{3}{4}}{2 \times 10}=\frac{15}{4} m)At the highest position there is only horizontal component of velocity.
$$\therefore \overrightarrow{v}=v_{x} \hat{i}=5 \hat{i} m / s$$
Q37. A body is projected horizontally with a velocity (3 m / s) and another with a speed of (30 m / s). The time taken by the first to reach the ground is ______ that taken by the second.
1) less
2) more
3) ((1 / 10)^{th})
4) same
Q38. A body is projected from the ground with a velocity (\vec{v}=(3 \hat{i}+10 \hat{j}) m / s). The maximum height attained and the range of the body respectively are (given (g=10 ms^{-2}))
1) 5 m and 6 m
2) 3 m and 10 m
3) 6 m and 5 m
4) 3 m and 5 m
Hint: $$H=\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{(u \sin \theta)^{2}}{2 g}=\frac{(10)^{2}}{2 \times 10}=5 m\\R=\frac{u^{2} \sin 2 \theta}{g}=\frac{2(u \sin \theta)(u \cos \theta)}{g}=\frac{2 \times 10 \times 3}{10}=6 m$$
Q39. Two projectiles A and B are thrown with the same speed but angles of (40^{\circ}) and (50^{\circ}) with the horizontal. Which projectile will fall earlier?
1) A
2) B
3) Both will fall at the same time
4) None of the above
Hint: (t=\frac{2 u \sin \theta}{g})Lesser is the value of (\theta), lesser is (\sin \theta) and hence lesser will be the time taken. Hence A will fall earlier
Q40. If (R) is the range of a projectile motion, then equation of its trajectory is
1) (y=x \cot \theta\left(1-\frac{x}{R}\right))
2) (y=x \cot \theta\left(1+\frac{x}{R}\right))
3) (y=x \tan \theta\left(1-\frac{x}{R}\right))
4) (y=x \tan \theta\left(1+\frac{x}{R}\right))
Q41. For a given velocity, a projectile has the same range R for two angles of projection. If (t_{1}) and (t_{2}) are the times of flight in the two cases, then
1) (t_{1} t_{2} \propto R^{2})
2) (t_{1} t_{2} \propto R)
3) (t_{1} t_{2} \propto \frac{1}{R})
4) (t_{1} t_{2} \propto \frac{1}{R^{2}})
Hint: For same range, angle of projection should be
(\theta) and
(90-\theta)(t_{1}=\frac{2 u \sin \theta}{g})(t_{2}=\frac{2 u \sin (90-\theta)}{g}=\frac{2 u \cos \theta}{g})$$t_{1} t_{2}=\frac{4 u^{2} \sin \theta \cos \theta}{g^{2}}=\frac{2}{g} \frac{\left(u^{2} \sin 2 \theta\right)}{g}=\frac{2 R}{g} \\ \therefore t_{1} t_{2} \propto R$$
Q42. If a particle is thrown at an angle with horizontal and time of flight T and range of projectile R are 10 second and 200 m , the velocity of projection is
1) (10 \sqrt{29} m / s)
2) (5 \sqrt{29} m / s)
3) (15 \sqrt{29} m / s)
4) (20 \sqrt{29} m / s)
Hint: (T=\frac{2 u \sin \theta}{g})(10=\frac{2}{10}(u \sin \theta))(u \sin \theta=50 m / s)(R=(u \cos \theta) T)(200=(u \cos \theta) \times 10)(u \cos \theta=50 m / s)$$u=\sqrt{(50)^{2}+(20)^{2}}=\sqrt{2900}=10 \sqrt{29} m / s$$
Q43. A ball rolls off the edge of a horizontal plane 4.9 m high. If it strikes the floor at a point 10 m horizontally away from the edge of the plane, speed of the ball at the instant it left the plane is
1) (10 ms^{-1})
2) (20 ms^{-1})
3) (30 ms^{-1})
4) (40 ms^{-1})
Hint: If (t) is the time of flight of the ball, then(t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 4.9}{9.8}}=1 s)Horizontal distance covered,(x=ut)(\therefore 10=u \times 1)(u=10 m / s)
Q44. A particle is projected at an angle of (60^{\circ}) above the horizontal with a speed of (10 m / s). After some time the direction of its velocity makes an angle of (30^{\circ}) above the horizontal. The speed of the particle at this instant is
1) (\frac{5}{\sqrt{3}} m / s)
2) (5 \sqrt{3} m / s)
3) (5 m / s)
4) (\frac{10}{\sqrt{3}} m / s)
Hint: For horizontal Motion,
(u \cos 60=v \cos 30)$$v=\frac{u \cos 60}{\cos 30}=\frac{10 \times \frac{1}{2}}{\sqrt{3} / 2}=\frac{10}{\sqrt{3}} m / s$$
Q45. The range of projectile projected at an angle (15^{\circ}) is (10 \sqrt{3} m). If it is fired with the same speed at angle of (30^{\circ}), then its range will be
1) 60 m
2) 45 m
3) 30 m
4) 15 m
Hint: (R=\frac{u^{2} \sin 2 \theta}{g})(\Rightarrow R \propto \sin 2 \theta)$$\therefore \frac{R_{1}}{R}=\frac{\sin 2 \theta_{1}}{\sin 2 \theta}=\frac{\sin 60}{\sin 30}=\sqrt{3}$$
$$R_{1}=\sqrt{3} R=\sqrt{3} \times 10 \sqrt{3}=30 m$$