Question

A boy throws a ball with speed u in a well of depth 14m as shown. on bounce with bottom of the well the speed of the ball gets halved. what should be the minimum value of u (in m/s) such that the ball may be able to reach

Answer 1

From the bottom of the well the the ball bounces up at a velocity u.

From kinematics equation :

V² = U² - 2gs

The g is negative since when the ball bounces back it experiences retardation.

V = final velocity

U = initial velocity

g = acceleration due to gravity = 10

S = the displacement = 14m

The final velocity = 0

Doing the substitution :

0 = u² - 2 × 10 × 14

0 = u² - 280

u² = 280

u = √280 = 16.73

The final speed at which the ball hits the bottom of the well is twice this :

V = 2 × 16.73 = 33.46

We want to get the initial velocity at which the ball is thrown.

We will use the following kinematics equation:

V² = U² + 2gs

33.46² = U² + 280

1119.57 - 280 = U²

839.57 = U²

U = √839.57 = 28.98 m/s