Paper Questions (Raw)
Q1. The greatest height reached by a projectile is ((1 / 4)^{\text{th}}) the horizontal range. The angle of projection is
1) (35^{\circ})
2) (45^{\circ})
3) (55^{\circ})
4) (25^{\circ})
Hint: According to given condition, (R=4H) (\frac{u^{2} \sin 2 \theta}{g}=\frac{4 u^{2} \sin ^{2} \theta}{2 g}) (2 \sin \theta \cos \theta=\frac{4 \sin ^{2} \theta}{2}) (\tan \theta=1) (\theta=45^{\circ})
Q2. The range of the particle when launched at an angle of (15^{\circ}) with the horizontal is 1.5 km . What is the range of the projectile when launched at an angle of (45^{\circ}) to the horizontal?
1) 1.5 km
2) 3.0 km
3) 6.0 km
4) 0.75 km
Hint: (\frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin 30}{g}) $$1.5=\frac{u^{2}}{2 g} \frac{u^{2}}{g}=3 R_{2}=\frac{u^{2} \sin 90}{g}=\frac{u^{2}}{g}=3 km$$
Q3. 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}(90^{\circ}-\theta)}{2 g}) $$ \therefore \frac{H_{1}}{H_{2}}=\frac{\sin ^{2} \theta}{\sin ^{2}(90^{\circ}-\theta)}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{\tan ^{2} \theta}{1} \therefore H_{1}: H_{2}:: \tan ^{2} \theta: 1 $$
Q4. A projectile has the maximum range of 500 m . If the projectile is now thrown up an inclined plane of (30^{\circ}) with the same speed, then the distance covered by it along the inclined plane will be
1) 250 m
2) 500 m
3) 750 m
4) 1000 m
Hint: (R=u^{2}(\frac{\sin 2 \theta}{g})=\frac{u^{2}}{g}) (( \because \theta=45^{\circ})) (\therefore 500=\frac{u^{2}}{g}) (u=\sqrt{(500 g)}) (v^{2}-u^{2}=2 gs) (0-(500 g)=2 \times(-g \sin 30^{\circ}) \times x) (x=500 m)
Q5. 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-gt=(10 \times \frac{\sqrt{3}}{2})-(10 \times \frac{1}{\sqrt{3}})=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)
Q6. An aeroplane moving horizontally with a speed of (720 km / h) drops a food pocket, while flying at a height of 396.9 m . The time taken by a food pocket to reach the ground and its horizontal range are ((Take g=9.8 m / s^{2}))
1) 3 s and 2000 m
2) 5 s and 500 m
3) 8 s and 1500 m
4) 9 s and 1800 m
Hint: (t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 3969}{9.8}}=9 s) (u=720 km / hr=720 \times \frac{5}{18}=200 m / s) (R=ut=200 \times 9=1800 m)
Q7. Two particles are projected in air with speed (u) at angles (\theta_{1}) and (\theta_{2}) (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then which one of the following is correct? where (T_{1}) and (T_{2}) are the time of flight.
1) ( \theta_{1}>\theta_{2})
2) ( \theta_{1}=\theta_{2})
3) ( T_{1}
4) ( T_{1}=T_{2})
Hint: (H_{1}>H_{2}) (\frac{u^{2} \sin ^{2} \theta_{1}}{2 g}>\frac{u^{2} \sin ^{2} \theta_{2}}{2 g}) (\therefore \theta_{1}>\theta_{2}) (T=\frac{2 u \sin \theta}{g} \Rightarrow T \propto \theta \Rightarrow T_{1}>T_{2})
Q8. A large number of bullets are fired in all directions with the same speed v. The maximum area on the ground on which these bullets will spread is
1) (\frac{\pi v^{2}}{g})
2) (\frac{\pi v^{4}}{g^{2}})
3) (\frac{\pi^{2} v^{4}}{g^{2}})
4) (\frac{\pi^{2} v^{2}}{g^{2}})
Hint: (R_{max}=\frac{v^{2} \sin 90^{\circ}}{g}=\frac{v^{2}}{g}) (Area =\pi(R_{max})^{2}=\frac{\pi v^{4}}{g^{2}})
Q9. A projectile is thrown into space so that it has the maximum possible horizontal range of 400 m . Taking the point of projection as the origin, the co-ordinate of the point where the velocity of the projectile is minimum are
1) ((400,200))
2) ((200,200))
3) ((400,100))
4) ((200,100))
Hint: Maximum horizontal range, (R=4H=400 m) (\Rightarrow H=100 m) Velocity of the projectile is minimum at highest position. ∴ Co-ordinate (( \frac{R}{2}, H))=((200,100))
Q10. A ball rolled off along the edge of a table (horizontal) with velocity (4 m / s). It hits the ground after 0.4 s , which one of the following statements is wrong?
1) The height of table is 0.8 m .
2) It hits the ground at an angle of nearly (60^{\circ}) with the vertical.
3) It covers a distance of 1.6 m from the edge of the table.
4) It hits the ground with a vertical velocity (4 m / s).
Hint: (h=ut-\frac{1}{2}gT^{2}=-\frac{1}{2}\times(-10)\times(0.4)^{2}=5\times 0.16=0.8 m) (\tan \theta=\frac{v_{y}}{v_{x}}=\frac{gt}{u_{x}}=\frac{10 \times 0.4}{4}=1 \Rightarrow \theta=45^{\circ}) (R=u_{x} T=4 \times 0.4=1.6 m) (v_{y}=-gT=-10 \times 0.4=-4 m / s)
Q11. Three balls of same masses are projected with equal speeds at angle (15^{\circ}, 45^{\circ}, 75^{\circ}) and their ranges are respectively (R_{1}, R_{2}) and (R_{3}) then
1) (R_{1}>R_{2}>R_{3})
2) (R_{1}
3) (R_{1}=R_{2}=R_{3})
4) (R_{1}=R_{3}
Hint: (R=\frac{u^{2} \sin 2 \theta}{g} \Rightarrow R \propto \sin 2 \theta) (\therefore R_{1}=R_{3}
Q12. A ball is thrown upwards and returns to the ground describing a parabolic path. Which of the following quantities remains constant?
1) Kinetic energy of the ball.
2) Speed of the ball.
3) Vertical component of velocity.
4) Horizontal component of velocity.
Q13. Two stones are projected with same velocity v at an angle (\theta) and ((90^{\circ}-\theta)). If H and (H_{1}) are greatest heights in the two paths, then (H_{1}+H=)
1) (\frac{v^{2}}{2 g})
2) (\frac{v^{2} \cos ^{2} \theta}{2 g})
3) (\frac{v^{2} \sin ^{2} \theta}{2 g})
4) (\frac{v^{2}}{4 g})
Hint: (H=\frac{v^{2} \sin ^{2} \theta}{2 g}) (H_{1}=\frac{v^{2} \sin ^{2}(90-\theta)}{2 g}=\frac{v^{2} \cos ^{2} \theta}{2 g}) (\therefore H_{1}+H=\frac{v^{2} \cos ^{2} \theta}{2 g}+\frac{v^{2} \sin ^{2} \theta}{2 g}=\frac{v^{2}}{2 g})
Q14. A projectile is given an initial velocity of ((\alpha \hat{i}+\beta \hat{j})) (m / s), where (\hat{i}) is along the ground and (\hat{j}) is along the vertical. The equation of its trajectory is
1) (y=\alpha x-\beta x^{2})
2) (y=\frac{\beta x}{\alpha}-\frac{gx^{2}}{2 \alpha^{2}})
3) (y=\frac{\alpha x}{\beta}-\frac{gx^{2}}{2 \alpha^{2}})
4) (y=\frac{\alpha x}{\beta}+\frac{gx^{2}}{2 \alpha^{2}})
Hint: (u=\alpha \hat{i}+\beta \hat{j}) (u=u_{x} \hat{i}+u_{y} \hat{j}) (u_{x}=u \cos \theta=\alpha) (u_{y}=u \sin \theta=\beta) Dividing eq. (ii) and (iii), (\tan \theta=\frac{\beta}{\alpha}) Equation of projectile is given by, (y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}=x \tan \theta-\frac{g x^{2}}{2(u \cos \theta)^{2}}=x(\frac{\beta}{\alpha})-\frac{g x^{2}}{2(\alpha)^{2}}=\frac{\beta x}{\alpha}-\frac{g x^{2}}{2 \alpha^{2}})
Q15. 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=2H) (\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))
Q16. If the velocity of projection, of a projectile, is increased by (2 \%) (other things remaining constant) the horizontal range will increase by
1) (1 \%)
2) (2 \%)
3) (4 \%)
4) (8 \%)
Hint: (R=\frac{u^{2} \sin 2 \theta}{g} \Rightarrow R \propto u^{2}) (\frac{dR}{R}=2 \times \frac{du}{u}=2 \times 2 \%=4 \%)
Q17. A body is projected at an angle of (30^{\circ}) to the horizontal with kinetic energy E. The kinetic energy at the topmost point is
1) zero
2) (\frac{E}{4})
3) (\frac{E}{2})
4) (\frac{3 E}{4})
Hint: (E=\frac{1}{2}mu^{2}) At highest position, K= 1 m (u cos 0) 2= E x cos2 30= 3E
Q18. A ball is dropped from a height (h) and another ball is projected horizontally from the same height. Then,
1) both will reach the ground at different times.
2) both will strike the ground will different speeds.
3) both will strike the ground with different vertical component of velocity.
4) both will strike the ground with same speed.
Hint: When the balls hit the ground, their vertical velocities are equal. However the horizontal velocities will be different. Hence the resultant velocities of the two balls are different, so the balls would hit the ground with different velocities. Both the balls would reach the ground at the same time because their initial vertical velocities, acceleration and distance covered are all equal.
Q19. If (range) (^{2}) is 48 times (maximum height) (^{2}), then angle of projection is
1) (45^{\circ})
2) (60^{\circ})
3) (75^{\circ})
4) (30^{\circ})
Hint: (R^{2}=48 H^{2}) $$ (\frac{u^{2} \sin 2 \theta}{g})^{2}=48(\frac{u^{2} \sin ^{2} \theta}{2 g})^{2} \frac{u^{4}(2 \sin \theta \cos \theta)^{2}}{g^{2}}=48(\frac{u^{4} \sin ^{4} \theta}{4 g^{2}}) $$
(16 \cos ^{2} \theta=48 \sin ^{2} \theta) (\tan ^{2} \theta=\frac{1}{3} \Rightarrow \tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ})
Q20. A bomb is dropped from an aeroplane moving horizontally at constant speed. When air resistance is taken into consideration, the bomb
1) falls to earth exactly below the aeroplane.
2) fall to earth behind the aeroplane.
3) falls to earth ahead of the aeroplane.
4) flies with the aeroplane.
Hint: Due to air resistance, horizontal velocity of bomb will decrease so it will fall behind the aeroplane.
Q21. The range of a projectile is R , when the angle of projection is (40^{\circ}). For the same velocity of projection and range, the other possible angle of projection is
1) (45^{\circ})
2) (50^{\circ})
3) (60^{\circ})
4) (40^{\circ})
Hint: For the same velocity of projection and range, the sum of the two angles of projection is (90^{\circ}). (\therefore 0=90^{\circ}-40^{\circ}=50^{\circ})
Q22. 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-θ. (h_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g}) (h_{2}=\frac{u^{2} \sin ^{2}(90^{\circ}-\theta)}{2 g}=\frac{u^{2} \cos ^{2} \theta}{2 g}) (R=u^{2}(\frac{\sin 2 \theta}{g})=\frac{2(u \sin \theta)(u \cos \theta)}{g}=4(\sqrt{\frac{u^{2} \sin ^{2} \theta}{2 g}})(\sqrt{\frac{u^{2} \cos ^{2} \theta}{2 g}})=4 \sqrt{h_{1} h_{2}})
Q23. A projectile is fired at an angle of projection (60^{\circ}), such that its range is 20 m . The equation of the projectile path will be
1) (20 y=\sqrt{3} x-\sqrt{3} x^{2})
2) (20 \sqrt{3} y=x-\sqrt{3} x^{2})
3) (20 y=20 \sqrt{3} x-\sqrt{3} x^{2})
4) (20 y=\sqrt{3} x-20 \sqrt{3} x^{2})
Hint: (y=x \tan \theta(1-\frac{x}{R})=x \tan 60(1-\frac{x}{R})=\sqrt{3} x(1-\frac{x}{20})=\sqrt{3} x-\frac{\sqrt{3} x^{2}}{20}) (\therefore 20 y=20 \sqrt{3} x-\sqrt{3} x^{2})
Q24. 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)
Q25. A shell fired from the ground is just able to cross in a horizontal direction, the top of a wall 90 m away and 45 m high. The velocity projection of the shell will be
1) (30 ms^{-1})
2) (30 \sqrt{2} ms^{-1})
3) (303m-1)
4) (15ms-1)
Hint: (H=\frac{u^{2} \sin ^{2} \theta}{2 g}) (R=\frac{u^{2} \sin 2 \theta}{g}=\frac{2 u^{2} \sin \theta \cos \theta}{g}) (\frac{H}{R}=\frac{1}{4} \tan \theta=\frac{45}{180}) (\tan \theta=1 \Rightarrow \theta=45^{\circ}) (v^{2}=u_{y}^{2}-2 gH=0) (u_{y}^{2}=2 \times 10 \times 45=900) (u_{y}=u \sin \theta=30 m / s) (u \sin 45=30) (u=30 \sqrt{2} m / s)
Q26. 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=) 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}})
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}=(\frac{2 u^{2}}{g})(\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}). The 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: Standard equation of projectile motion
(y=(\tan \theta) x-\frac{gx^{2}}{2 u^{2} \cos ^{2} \theta}=\sqrt{3} x-x^{2}) (\tan \theta=\frac{\sin \theta}{\cos \theta}=\sqrt{3}) (\therefore \sin \theta=\sqrt{3} \cos \theta) (\frac{2 u^{2} \cos ^{2} \theta}{g}=1 \Rightarrow \frac{u^{2} \cos ^{2} \theta}{g}=\frac{1}{2}) $$ H=\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{u^{2}(3 \cos ^{2} \theta)}{2 g}=\frac{3}{4}=0.75 m $$
Q29. The equation of a projectile path is given by, (y=\sqrt{3} x-x^{2}). The 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: Standard equation of projectile motion (y=(\tan \theta) x-\frac{gx^{2}}{2 u^{2} \cos ^{2} \theta}=\sqrt{3} x-x^{2}) (\frac{2 u^{2} \cos ^{2} \theta}{g}=1) (\therefore v=u \cos \theta=\sqrt{\frac{g}{2}}=\sqrt{5} m / s)
Q30. The instantaneous height of a projectile and the distance x covered by it are as follows: (x=5 t), (y=5 t-2 t^{2}). The acceleration due to gravity, in SI units, is
1) (2 m / s^{2})
2) (4 m / s^{2})
3) (6 m / s^{2})
4) (8 m / s^{2})
Hint: x = (u cos 0) t=5t m u cos θ=5 y =(usin θ)t - 1 a ay t2 = 5t - 2t 2 u }sin0= (\frac{1}{2}\textrm{a}_{y}=2\Rightarrow\textrm{a}_{y}=4\textrm{m}/\textrm{s}^{2})