Question

A tennis ball is dropped from 1.95 m above the ground. It
rebounds to a height of 0.932 m.

Part 1.With what velocity does it hit the ground? The acceleration
of gravity is 9.8 m/s 2 . (Let down be negative.) Answer in units
of m/s.

Part 2. With what velocity does it leave the ground? Answer in units of m/s.

Part 3. If the tennis ball were in contact with the ground for 0.00859 s, find the acceleration given to the tennis ball by the ground. Answer in units of m/s 2 .

Answer #1

here,

the height from which the ball is dropped , h1 = 1.95 m

the height to which it rebounds , h2 = 0.932 m

a)

the velocity of ball before it hits the ground , u = - sqrt(2 * g * h1)

u = - sqrt(2 * 9.81 * 1.95) m/s

u = - 6.19 m/s

b)

the velocity of ball after it leaves the ground , v = sqrt(2 * g * h2)

v = sqrt(2*9.81*0.932) = 4.28 m/s

c)

time taken , t = 0.00859 s

let the acceleration of ball be a

using first equation of motion

v = u + a * t

4.28 = - 6.19 + a * 0.00859

solving for a

a = 1.22 * 10^3 m/s^2the acceleration of ball is 1.22 * 10^3 m/s^2

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