J J Science Academy
1) 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
Hint: At $t=1 s$ $$v_{X}=\frac{d x}{d t}=3 c t^{2}=3 c \\\\v_{Y}=\frac{d y}{d t}=3 b t^{2}=3 b \\\\ \therefore v_{net }=\sqrt{v_{X}^{2}+v_{Y}^{2}}=\sqrt{(3 c)^{2}+(3 b)^{2}}=3 \sqrt{c^{2}+b^{2}} m / s$$
2) 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 $\theta=45^{\circ}$
3) On which of the following, is the maximum height achieved by a projectile, independent?
4) The angle of projection for same maximum range is
5) If $R$ is the range of a projectile motion, then equation of its trajectory is
6) In a projectile motion, the velocity
7) 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 $R=$
Hint: Same range is obtained for an angles $\theta$ and $(90-\theta)$. $$h_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g} \\\\h_{2}=\frac{u^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 g}=\frac{u^{2} \cos ^{2} \theta}{2 g} \\\\R=u^{2}\left(\frac{\sin 2 \theta}{g}\right)=\frac{2(u \sin \theta)(u \cos \theta)}{g}=4\left(\sqrt{\frac{u^{2} \sin ^{2} \theta}{2 g}}\right)\left(\sqrt{\frac{u^{2} \cos ^{2} \theta}{2 g}}\right)=4 \sqrt{h_{1} h_{2}}$$
8) A particle is projected with a velocity v so that its horizontal range is twice the greatest height attained. The horizontal range is
Hint: $R=2 H$ $$\frac{v^{2} \sin 2 \theta}{g}=2 \times \frac{v^{2} \sin ^{2} \theta}{2 g}\\\\2 \sin \theta \cos \theta=\sin ^{2} \theta\\\\ \tan \theta=2\\\\ \therefore \sin \theta=\frac{2}{\sqrt{5}} and \cos \theta=\frac{1}{\sqrt{5}}\\\\R=\frac{v^{2} \times \sin 2 \theta}{g}=v^{2} \times \frac{2 \sin \theta \cos \theta}{g}=v^{2} \times \frac{2}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}\\\\ \therefore R=\frac{4 v^{2}}{5 g}$$
9) 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
Hint: $$y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}=40 \tan 45^{\circ}-\frac{10(40)^{2}}{2(40)^{2} \cos ^{2} 45^{\circ}}=40-10=30 m$$
10) 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?
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}$$
11) For a projectile motion, if $x=8 t$ and $y=2 t-3 t^{2}$, then its time of flight is
Hint: For maximum height,$v_{y}=2-6 t=0$$t=\frac{1}{3} s=T_{a}$$T=2 T_{a}=\frac{2}{3} s$
12) A shell fired from a canon can cover maximum horizontal distance of 10 km . Then velocity of projection is
Hint: $R_{\max }=\frac{u^{2}}{g}=10000$$u^{2}=9.8 \times 10000$$u^{2}=98000$$u=\sqrt{98000} m / s$
13) 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
Hint: The horizontal momentum does not change.The change in vertical momentum is$\Delta p=mv \sin 45-(-mv \sin 45)=\frac{2 mv}{\sqrt{2}}=\sqrt{2} mv$
14) 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 $v_{A}=(\tan 37=\frac{3}{4})$
Hint: To hit the bird simultaneously, the vertical component must be same.$\therefore v_{A} \cos 53=v_{B} \cos 37$ $$v_{A}=v_{B} \frac{\cos 37}{\cos 53}=v_{B} \frac{\cos 37}{\cos (90-53)}=v_{B} \frac{\cos 37}{\sin 37}=\frac{v_{B}}{\tan 37}=60 \times \frac{4}{3}=80 m / s$$
15) 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 $g=10 m / s^{2})$
Hint: $$H=\frac{v^{2} \sin ^{2} \theta}{2 g}=\frac{(20 \sqrt{2})^{2} \sin ^{2} 45}{2 \times 10}=\frac{800 \times \frac{1}{2}}{2 \times 10}=20 m\\\\R=\frac{v^{2} \sin 2 \theta}{g}=\frac{(20 \sqrt{2})^{2} \sin (2 \times 45)}{10}=\frac{800 \times 1}{10}=80 m\\\\ Displacement =\sqrt{H^{2}+\left(\frac{R}{2}\right)^{2}}=\sqrt{(20)^{2}+(40)^{2}}=\sqrt{400+1600}=\sqrt{2000}=20 \sqrt{5} m\\\\v_{av}=\frac{ Displacement }{ Time }=\frac{20 \sqrt{5}}{2}=10 \sqrt{5} m / s$$
16) 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?
Hint: $H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g} \Rightarrow H \propto u^{2} \sin ^{2} \theta= same$ $$\frac{u_{1}}{u_{2}}=\frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{\sin 60}{\sin 45}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$$
17) The angle of projection, for which the horizontal range and the maximum height of a projectile are in the ratio of $2: 1$ is
Hint: According to given condition, $R=2 H$ $$\frac{u^{2} \sin 2 \theta}{g}=\frac{2 u^{2} \sin ^{2} \theta}{2 g}\\\\2 \sin \theta \cos \theta=\frac{2 \sin ^{2} \theta}{2}\\\\ \tan \theta =2 \\\\ \theta =\tan ^{-1}(2)$$
18) 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
Hint: $y=(\tan \theta) x-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}=\frac{x}{\sqrt{3}}-\frac{1}{2} g x^{2}$ $$\tan \theta =\frac{1}{\sqrt{3}} \\\\ \theta =30^{\circ}$$
19) A body is projected with initial velocity of $(8 \hat{i}+6 \hat{j}) m / s$. The horizontal range is
Hint: $$R=\frac{u^{2} \sin 2 \theta}{g}=\frac{2(u \sin \theta)(u \cos \theta)}{g}=\frac{2 \times 8 \times 6}{10}=9.6 m$$
20) 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?
Hint: $$R_{1}=\frac{u^{2} \sin 2 \theta_{1}}{g} \\\\150=\frac{u^{2} \sin 30}{g} \\\\ \frac{u^{2}}{g}=300 \\\\R_{2}=\frac{u^{2} \sin 2 \theta_{2}}{g}=\frac{u^{2} \sin 90}{g}=300 m$$
21) 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
Hint: Maximum height for mass $m_{1}, H_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g}$Maximum height for $m_{2}, H_{2}=\frac{u^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 g}$ $$\therefore \frac{H_{1}}{H_{2}}=\frac{\sin ^{2} \theta}{\sin ^{2}\left(90^{\circ}-\theta\right)}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{\tan ^{2} \theta}{1} \\\\ \therefore H_{1}: H_{2}:: \tan ^{2} \theta: 1$$
22) 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
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}$$
23) A stone is projected from the ground with velocity $50 m / s$ at an angle of $30^{\circ}$. It crosses a wall after 3 s . How far beyond the wall the stone will strike the ground? [ $g=10 m / s^{2})$
Hint: $T=\frac{2 u \sin \theta}{g}=\frac{2 \times 50 \times \frac{1}{2}}{10}=5 s$Time to cross the wall $(=3 s)$ (given)Time in air after crossing the wall $(=(5-3)=2 s)$∴ Distance travelled beyond the wall $$x=(u \cos \theta) t=50 \times \frac{\sqrt{3}}{2} \times 2=86.6 m$$
24) Two projectiles, one fired from the surface of the earth with speed $5 m / s$ and other fired from the surface of a planet with initial speed of $3 m / s$, traces identical trajectories. Neglecting friction effect, the value of acceleration due to gravity on the planet is
Hint: Here, $\frac{u_{e}^{2}}{g_{e}}=\frac{u_{p}^{2}}{g_{p}} \quad(Range = constant)$ $$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}$$
25) From the top of a tower, three balls $A, B$ and C whose mass is in the ratio $1: 2: 3$ are thrown with the same speed $v_{0}$ as shown. The ratio of speed with which they strike the ground is
Hint: $K_{1}+U_{1}=K_{2}+U_{2}$$\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} and h are same for all three cases, v_{A}=v_{B}=v_{C}$
26) A body, of mass 4 kg , is fired with velocity $(5 \hat{i}+4 \hat{j}) ms^{-1}$. Its minimum kinetic energy will be
Hint: At highest position, we get minimum kinetic energy with velocity $u \cos \theta=5 m / s.$ $$K_{\min }=\frac{1}{2} m(u \cos \theta)^{2}=\frac{1}{2} \times 4 \times 25=50 J$$
27) The equation of a projectile path is given by, $y=\sqrt{3} x-x^{2}$. The range of motion is
Hint: Standard equation of projectile motion $$y=(\tan \theta) x-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}=\sqrt{3} x-x^{2}$$ $$\frac{2 u^{2} \cos ^{2} \theta}{g}=1$$ $\tan \theta=\sqrt{3}$ $$R=\frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2}(2 \sin \theta \cos \theta)}{g}=\left(\frac{2 u^{2}}{g}\right)(\sin \theta \cos \theta)=\frac{\sin \theta \cos \theta}{\cos ^{2} \theta}=\tan \theta=\sqrt{3} m$$
28) The equation of a projectile path is given by, $y=\sqrt{3} x-x^{2}$. Maximum height reached by the body is
Hint: $$H=\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{u^{2}\left(3 \cos ^{2} \theta\right)}{2 g}=\frac{3}{4}=0.75 m$$
29) The equation of a projectile path is given by, $y=\sqrt{3} x-x^{2}$. The angle of projection is
Hint: $\tan \theta = \sqrt{3} \Rightarrow \theta = 60^{\circ}$
30) The equation of a projectile path is given by, $y=\sqrt{3} x-x^{2}$. The velocity at the highest point is
Hint: $$v=u \cos \theta=\sqrt{\frac{g}{2}}=\sqrt{5} m / s$$
31) The equation of a projectile path is given by, $y=\sqrt{3} x-x^{2}$. The velocity of projection is
Hint: $$u=\frac{\sqrt{5}}{\cos 60}=2 \sqrt{5} m / s$$
32) 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
Hint: $R=\frac{u^{2} \sin 2 \theta}{g} \Rightarrow R \propto \sin 2 \theta$$\therefore X=Z
33) Three projectiles are fired at an angle of projection of $20^{\circ}, 45^{\circ}, 70^{\circ}$, with same velocity. If the maximum heights reached by the projectiles are $H_{1}, H_{2}$ and $H_{3}$, then $H_{1}+H_{3}=$
Hint: $H_{2}=\frac{u^{2} \sin ^{2} 45}{2 g}=\frac{u^{2}}{4 g}$$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}$$
34) Three projectiles are fired at an angle of projection of $20^{\circ}, 45^{\circ}, 70^{\circ}$, with same velocity. If the time of flights of the projectiles are $T_{1}, T_{2}$ and $T_{3}$, then $T_{1}^{2}+T_{3}^{2}=$
Hint: $T_{2}=\frac{2 u \sin \theta_{2}}{g}$$\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}$ $$T_{1}^{2}+T_{3}^{2}=\frac{4 u^{2} \sin ^{2} \theta_{1}}{g}+\frac{4 u^{2} \cos ^{2} \theta_{1}}{g}=\frac{4 u^{2}}{g}=2 T_{2}^{2}$$
35) 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 $(g=10 m / s^{2})$
Hint: $T=\frac{2 u \sin \theta}{g}=\frac{2 \times 10 \times \frac{\sqrt{3}}{2}}{10}=\sqrt{3} s$ $$v_{x}=u \cos 60=10 \times 0.5=5 m / s \\\\v_{y}=u \sin 60-g t=\left(10 \times \frac{\sqrt{3}}{2}\right)-\left(10 \times \frac{1}{\sqrt{3}}\right)=5 \sqrt{3}-\frac{10}{\sqrt{3}}=\frac{5}{\sqrt{3}}=\frac{5 \sqrt{3}}{3} m / s \\\\ \vec{v}=v_{x} \hat{i}+v_{y} \hat{j}=5 \hat{i}+\frac{5 \sqrt{3}}{3} \hat{j} m / s$$
36) 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
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. $$\therefore \overrightarrow{v}=v_{x} \hat{i}=5 \hat{i} m / s$$
37) 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.
38) 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 $g=10 ms^{-2})$
Hint: $$H=\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{(u \sin \theta)^{2}}{2 g}=\frac{(10)^{2}}{2 \times 10}=5 m\\\\R=\frac{u^{2} \sin 2 \theta}{g}=\frac{2(u \sin \theta)(u \cos \theta)}{g}=\frac{2 \times 10 \times 3}{10}=6 m$$
39) 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?
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
40) If $R$ is the range of a projectile motion, then equation of its trajectory is
41) 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
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$$
42) If a particle is thrown at an angle with horizontal and time of flight T and range of projectile R are 10 second and 200 m , the velocity of projection is
Hint: $T=\frac{2 u \sin \theta}{g}$$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$ $$u=\sqrt{(50)^{2}+(20)^{2}}=\sqrt{2900}=10 \sqrt{29} m / s$$
43) 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
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$
44) 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
Hint: For horizontal Motion,$u \cos 60=v \cos 30$ $$v=\frac{u \cos 60}{\cos 30}=\frac{10 \times \frac{1}{2}}{\sqrt{3} / 2}=\frac{10}{\sqrt{3}} m / s$$
45) 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
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$$