Question

A tennis ball is dropped from 1.3 m above

Answer 1

Answer:

2.65m/s

Explanation:

Using the equation of motion:

v² = u²+2a∆S where

v is the final velocity

u is the initial velocity

∆S is the change in distance

a is the acceleration

Given

u = 0m/s

a = 9.8m/s²

∆S = 1.3-0.943

∆S = 0.357m

Substituting the given parameters into the formula

v² = 0²+2(9.8)(0.357)

v² = 0+6.9972

v² = 6.9972

v=√6.9972

v = 2.65m/s

Hence the velocity at which it hit the ground is 2.65m/s