Question

3. A boy throws a ball with speed u in a well of depth 14 mas shown. Onbounce with bottom of the
speed of the ball gets halved. What should be the minimum value of u (inm/s) such that the ball may be at
to reach his hand again? It is given that his hands are at 1 m height from top of the well while throwing and catching
​

Answer 1

Explanation:

The ball needs to travel 15m (Depth 14m and Boy at 1m), to reach the boy’s hand.

What should be the initial speed so as to counteract gravity and travel 15m.

Let’s take the minimum speed required at the bottom of the well be x and gravity be 10m/s^2

so in some time t sec, the velocity of the ball x m/s, becomes zero at acceleration -10m/s^2 (Retardation)

a = (v-u)/t

-10m/s^2 = (0-x)/t

==>x = 10t or t = x/10

So, the ball has to travel 15m in time t sec, acceleration of -10m/s^2 and initial velocity of x m/s

and using

s = ut + (1/2)(at^2)

15 = xt + (1/2)(-10)(t^2)

==> 15 = (1/10)x^2 - (1/20)(x^2)

==> x^2 = 300

==> x = 17.3205 m/s

So Speed at the bottom of the well is twice that, and hence u1 = 34.64 m/s

And if speed at bottom of well is 34.64m/s,

It had been accelerating at 10m/s^2 for the distance of 15 m

Hence

10 = (34.64-u)/t ==> 10t = 34.64 - u ==>u = 34.64 - 10t

So,

15 = ut + 1/2 (10)(t^2)

==> 15 = t(34.64 - 10t) +5t^2 = 34.64t - 10t^2 + 5t^2

==> 5t^2 -34.64t +15 = 0

Which gives 0.4641 sec

==> u = 34.64 - 10*(0.4641) = 30 m/s

Good Luck

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