## Drops of water fall from the root of a building 9 m high at regular intervals of time, the first drop…

Q: Drops of water fall from the root of a building 9 m high at regular intervals of time, the first drop reaching the ground at the same instant fourth drop starts its fall. What are the distances of the second and third drops from the roof ? (g = 10 ms )

(a) 6 m and 2 m

(b) 6 m and 3 m

(c) 4 m and 1 m

(d) 4 m and 2 m

Ans:(c)

Sol: Time taken by first drop to reach the ground $\displaystyle t = \sqrt{\frac{2 h}{g}}$ $\displaystyle t = \sqrt{\frac{2 \times 9}{10}}$

t = 1.36 sec

Time interval between successive drops
= 1.36/3 sec

Applying , $\displaystyle S = u t + \frac{1}{2}a t^2$

Distance of 2rd drop, $\displaystyle S_2 = 0 + \frac{1}{2}(10) (\frac{2}{3}\times 1.36)^2$

S2 = 4 m

Distance of 3rd drop, $\displaystyle S_3 = 0 + \frac{1}{2}(10) (\frac{1}{3}\times 1.36)^2$

S3 = 1 m

## A ball is dropped from top of a building. The ball takes 0.5 s to fall past the 3 m length of a window…

Q: A ball is dropped from top of a building. The ball takes 0.5 s to fall past the 3 m length of a window some distance from top of building. With what speed does the ball pass the top of window ?

(a) 6 m/s

(b) 12 m/s

(c) 7 m/s

(d) 3.5 m/s

Ans:(d)

Sol: Let x be the distance of top of window from top of the building, and let the ball take t sec to reach there.

Applying , $\displaystyle s = u t + \frac{1}{2} a t^2$ $\displaystyle x = 0 + \frac{1}{2} (10) t^2$

x = 5 t 2 …(i)

Now , $\displaystyle x + 3 = 0 + \frac{1}{2} (10) (t + 0.5)^2$ $\displaystyle x + 3 = 5(t^2 + t + 1/4)$ …(ii)

Subtract (i) from (ii)

3=5(1/4+t)

3 =5/4 + 5t

3-5/4 = 5t

7/4 = 5t

or t = 7/20 sec

From, v = u + at

v = 0 + 10× 7/20

= 3.5 m/s

## A ball is thrown upwards with a speed V from a height ‘h’ above the ground. The time taken…

Q: A ball is thrown upwards with a speed V from a height ‘h’ above the ground. The time taken by the ball to hit the ground is

(a) $\displaystyle \sqrt{\frac{2 h}{g}}$

(b) $\displaystyle \sqrt{\frac{8 h}{g}}$

(c) $\displaystyle \frac{\sqrt{V^2 + 2 g h}}{g} + \frac{V}{g}$

(d) $\displaystyle \sqrt{\frac{2 h}{g}} + \frac{V}{g}$

Ans:(c)

Sol : Time taken to reach highest point from the height h is obtained from

v = u + at

0 = V – gt,

t = V/g …(i)

Now , Height attained is obtained from $\displaystyle v^2 = u^2 + 2 a s$ $\displaystyle 0 = V^2 - 2 g s$ $\displaystyle s = \frac{V^2}{2 g}$

Total height for downward journey is $\displaystyle = \frac{V^2}{2 g} + h$

Time taken to hit the ground is obtained from $\displaystyle s = ut + \frac{1}{2}a t^2$ $\displaystyle \frac{V^2}{2 g} + h = 0 + \frac{1}{2}g t^2$ $\displaystyle t = \frac{\sqrt{V^2 + 2 g h}}{g}$

Total time taken $\displaystyle = \frac{\sqrt{V^2 + 2 g h}}{g} + \frac{V}{g}$

## A balloon is going upwards with velocity 12 m/sec. It release a packet when it is at a height of …

Q: A balloon is going upwards with velocity 12 m/sec. It release a packet when it is at a height of 65 m from the ground. How much time the packet will take to reach the ground if g = 10 m/s2

(a) 5 sec

(b) 6 sec

(c) 7 sec

(d) 8 sec

Ans:(a)

Sol: As packet is to fall to the ground and it is released initially with the velocity of balloon, so

u =-12 m/s, $\displaystyle s = ut + \frac{1}{2}at^2$ $\displaystyle 65 = -12t + \frac{1}{2}(10)t^2$ $\displaystyle 65 = -12t + 5 t^2$ $\displaystyle 5 t^2 -12 t - 65 = 0$ $\displaystyle t = \frac{12 \pm \sqrt{144 - 4(5)(-65)}}{2\times 5}$ $\displaystyle t = \frac{12\pm \sqrt{144 + 1300}}{10}$ $\displaystyle t = \frac{12 \pm 38}{10}$

t = 5 sec

## A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8 s …

Q: A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8 s, stone is released from the balloon. The stone will (g = 10 m/s2 )

(a) cover a distance of 40 m

(b) have a displacement of 50 m

(c) reach the ground in 4 s

(d) begin to move down after being released

Ans: (c)

Sol: Taking vertical upward motion of balloon for 8 sec.

v = u + a t

= 0 + 1.25 × 8

= 10 m/s $\displaystyle s = ut + \frac{1}{2}a t^2$ $\displaystyle s = 0 + \frac{1}{2}\times 1.25 \times 8^2$

= 40 m

Taking vertical downward motion of stone released from balloon, we have $\displaystyle s = ut + \frac{1}{2}a t^2$ $\displaystyle 40 = -10 t + \frac{1}{2}(10) t^2$ $\displaystyle 5 t^2 -10 t -40 = 0$ $\displaystyle t^2 - 2 t - 8 = 0$

On solving, t = 4 s