Question

10) A bowling ball released from rest at the top of a 200 m building falls to the ground under the influ-

Answer 1

(a) The speed at 5 s is 49 m/s.

(b) The distance travelled by the bowling ball at 32 m/s is 52.24m.

Explanation:

As the ball is released from the top of the building, it will be exhibiting free fall. So the acceleration due to gravity will be acting upon the bowling ball.

As per the equations of motion, the velocity can be determined for a particular time if we know the initial speed, acceleration and time.

With the help of first equation of motion,

$v = u +at$

So u =0, a = 9.8 m/$s^{2}$, t = 5 s

So, v = 0 + (9.8×5)=49 m/s

Thus, the speed at 5 s is 49 m/s.

When it is moving at a speed of 32 m/s, then the distance travelled by the ball is

$2as = v^{2}-u^{2}$

Since u =0 and v = 32 m/s with a = 9.8, then

$s=\frac{v^{2}-u^{2}}{2 a}=\frac{32^{2}-0^{2}}{2 \times 9.8}=\frac{1024}{19.6}=52.24 \mathrm{m}$

So the distance travelled is 52.24 m.