J J Science Academy
1. A ball is dropped from a bridge 125 meter above a river. After the ball has been falling for two second, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time? $(g = 10 m / s^2)$
Hint: For first ball, $(s = \frac{1}{2} gt^2)$ $$125 &= \frac{1}{2} \times 10 \times t^2 \\ t &= 5 s$$ For second ball, $(t_1 = 3 s)$ $(s_1 = ut_1 + \frac{1}{2} gt_1^2)$ $(125 = 3u + \frac{1}{2} \times 10 \times 9)$ $(u = \frac{80}{3} = 26.7 m / s)$
2. A ball is dropped from a bridge 125 meter above a river. After the ball has been falling for two second, a second ball is thrown straight down after it. what is the velocity of the first ball when it hits the water? $(g = 10 m / s^2)$
Hint: v = gt = 10 x 5 = 50 m/s
3. A ball is dropped from a bridge 125 meter above a river. After the ball has been falling for two second, a second ball is thrown straight down after it. what is the velocity of the second ball? $(g = 10 m / s^2)$
Hint: $(v = u + gt = 26.7 + (10 \times 3) = 26.7 + 30 = 56.7 m / s)$
4. Two bodies of different masses $(m_a)$ and $(m_b)$ are dropped from two different heights, a and b . The ratio of time taken by the two to drop through these distances is
Hint: $(s = \frac{1}{2} gt^2 \Rightarrow t = \sqrt{\frac{2 s}{g}} \Rightarrow t \propto \sqrt{s})$ $(\frac{t_a}{t_b} = \frac{\sqrt{a}}{\sqrt{b}})$
5. A body is fired vertically upwards, with a velocity $(60 m / s)$. The time intervals between the instants when it has a vertical displacement of 120 m , is
Hint: $(h = ut - \frac{1}{2} gt^2)$ $(120 = 60t - 5t^2)$ $(t^2 - 12t + 24 = 0)$ $(t^2 - 12t = -24)$ $(t^2 - 12 t + 36 = -24 + 36)$ $((t - 6)^2 = 12)$ $(t = 6 \pm \sqrt{12})$ $(t_1 = 6 - \sqrt{12} t_2 = 6 + \sqrt{12})$ $(\Delta t = t_2 - t_1 = 2 \sqrt{12} = 4 \sqrt{3} s)$
6. A body is fired vertically upwards, with a velocity $(60 m / s)$, the change in velocity of the body after a time interval of 5 s is
Hint: $(v = u - gt = 60 - 50 = 10 m / s)$ $(\Delta v = v_2 - v_1 = 10 - 60 = -50 m / s)$
7. A body dropped from the top of a tower covers a distance 7 x in the last second of its journey where x is the distance covered in first second. How much time does it take to reach the ground?
Hint: In first second distance travelled, $(x = \frac{1}{2} gt^2 = \frac{1}{2} \times 10 \times (1)^2 = 5 m)$ In last second, $(7x = 35 m)$ $(s_n = u + \frac{g}{2}(2n - 1))$ $(35 = 0 + 5(2n - 1))$ $(2n - 1 = 7)$ $(n = 4 s)$
8. A body, fired vertically upwards with velocity u, attains the same speed after time T. At two instants, $(t_1)$ and $(t_2)$, it crosses a point in its path. The $(t_1 + t_2)$ =
Hint: $(h = ut - \frac{1}{2} gt^2)$ $(2h = 2ut - gt^2)$ $(gt^2 - 2ut + 2h = 0)$ $(t_{1,2} = \frac{ \pm 2u + \sqrt{4u^2 - 4g(2h)}}{2g} = \frac{2u \pm \sqrt{4u^2 - 8gh}}{2g})$ $(t_1 = \frac{2u - \sqrt{4u^2 - 8gh}}{2g})$ $(t_2 = \frac{2u + \sqrt{4u^2 - 8gh}}{2g})$ $(t_1 + t_2 = 2(\frac{2u}{2g}) = \frac{2u}{g} = T)$
9. A player throws a ball vertically upwards with velocity u. At highest point,
Hint: At the highest point velocity of the ball becomes zero, but its acceleration is equal to g.
10. A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first five seconds of its motion. The height from which stone is dropped is
Hint: $(\frac{1}{2} g(5)^2 = \frac{1}{2} g[2t - 1])$ $(2t - 1 = 25)$ $(t = 13 s)$ $(h = \frac{1}{2} g(13)^2 = \frac{1}{2} \times 10 \times 169 = 5 \times 169 = 845 m)$
11. If the velocity of a body, thrown upwards, is increased by $(10 \%)$, the maximum height reached by it will increase by
Hint: $(v = u + 10\%u = 11u)$ $(v^2 \propto h)$ $(\frac{h_2}{h_1} = (\frac{v}{u})^2 = (\frac{11}{10})^2 = \frac{121}{100})$ $(\frac{h_2 - h_1}{h_1} \times 100 = (\frac{121 - 100}{100}) \times 100 = 21%)$
12. Two balls A and B of same masses are thrown from the top of the building. ' A ', thrown upwards with velocity v and 'B', thrown downwards with velocity v , then
Hint: $(v^2 = u^2 + 2gh)$ $(\therefore v = \sqrt{u^2 + 2gh})$ So. For both the cases, velocity will be equal.
13. An elevator car, whose floor to ceiling distance is equal to 2.7 m , starts ascending with constant acceleration of $(1.2 m s^-2)$. Two sec after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is
Hint: $(t = \sqrt{\frac{2h}{(g + a)}} = \sqrt{\frac{2 \times 2.7}{(9.8 + 1.2)}} = \sqrt{\frac{5.4}{11}} = \sqrt{0.49} = 0.7 s)$ As $(u = 0)$ and lift is moving upward with acceleration.
14. A ball is dropped from a bridge of 122.5 metre above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. Initial velocity of second ball, so that both hit the water at the same time is
Hint: For the first ball: $(H = \frac{1}{2} gt^2)$ $(22.5 = \frac{1}{2} gt^2)$ From eq. (i), $(t^2 = \frac{2 \times 122.5}{9.8} = 25)$ $(t = 5 s)$ For second ball: Time taken $(t = 2 s)$ $$122.5=u(t - 2) + \frac{1}{2}g(t - 2)^2$$ $(\therefore \frac{1}{2} gt^2 = ut - 2u + \frac{1}{2} gt^2 + 2g - 2gt)$ $(u(t - 2) = 2g(t - 1))$ From eq. (iii), $(3u = 8 \times 9.8 = 78.4)$ $(u = \frac{78.4}{3} = 26.1 m / s)$
15. A body thrown vertically up to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is
Hint: Time required to reach maximum height $(= t)$ Time required to reach half of maximum height is $(t_1)$ For downwards motion, $(h = \frac{1}{2} gt^2)$ $(h_1 = \frac{1}{2} gt_1^2 = \frac{h}{2})$ $(\therefore t_1^2 = \frac{t^2}{2} \Rightarrow t_1 = \frac{t}{\sqrt{2}})$ Total time $( = t + t_1 = t + \frac{t}{\sqrt{2}} = (1 + \frac{1}{\sqrt{2}}) t)$
16. Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion, with
Hint: Motion under gravity is independent of the mass of the body.
17. A body, of mass M , is thrown upwards with, from ground, with velocity u . The average acceleration of the body, during the time t in air is
18. A body rolls down a stair case of 5 steps. Each step has height 0.1 m and width 0.1 m . With what velocity will the body reach the bottom?
Hint: Horizontal distance covered by the body $( = 5 \times 0.1 = 0.5 m)$ Vertical distance covered by the body $( = 5 \times 0.1 = 0.5 m)$ If the time taken to hit the edge of $(5^{th})$ step is t,then $(ut = 0.5 \Rightarrow t = \frac{0.5}{u})$ $(\frac{1}{2} gt^2 = 0.5)$ $(\frac{1}{2} \times 10 \times \frac{0.25}{u^2} = 0.5)$ $(u^2 = \frac{10}{4} = \frac{5}{2})$ $(u = \sqrt{\frac{5}{2}} m / s)$
19. A player throws a ball vertically upwards with velocity u. At highest point,
Hint: At the highest point velocity of the ball becomes zero, but its acceleration is equal to g.
20. Water drops fall from the roof of a building 20 m high at regular intervals of time. The first drop reaching the ground at the same instant fifth drop starts its fall. The time between two successive drops is $(g = 10 m / s^2)$
Hint: Time taken by first drop to reach the ground $(t' = \sqrt{\frac{2s}{g}} = \sqrt{\frac{2 \times 20}{10}} = 2 s)$ Time interval between successive drops, $(t = \frac{t'}{4} = \frac{2}{4} = 0.5 s)$
21. If the velocity of a body, thrown upwards, is increased by $(10 \%)$, the maximum height reached by it will increase by
Hint: $(v = u + 10\% u = \frac{11}{10} u)$ $(v^2 \propto h)$ $(\frac{h_2}{h_1} = (\frac{v}{u})^2 = (\frac{11}{10})^2 = \frac{121}{100})$ $(\frac{h_2 - h_1}{h_1} \times 100 = (\frac{121 - 100}{100}) \times 100 = 21%)$
22. A body is dropped from the top of a tower, 30 m high. At the same time a body is fired from the ground, vertically upward, along the same line, with a speed of $(10 m / s)$. The time after which they will collide with each other is
Hint: Two bodies are collided at height h after time t. $$h = 10t - \frac{1}{2}gt^2$$ $$H - h = \frac{1}{2}gt^2$$ From eqs. (i) and (ii), $$H - 10t + \frac{1}{2}gt^2 = \frac{1}{2}gt^2$$ $$t = \frac{H}{10} = \frac{30}{10} = 3 s$$
23. A ball thrown up from the ground is observed at a certain height after times 4 s and 6 s . If the acceleration due to gravity is $(10 m s^-2)$, the velocity of throw is
Hint: Time taken by the ball to reach upto highest point $(t = 4 + (\frac{6 - 4}{2}) = 5 s)$ $(v = u + gt = 0 + (10 \times 5) = 50 m / s)$