Question

A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?

Answer 1

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration $g=9.8 m/s^2$ towards the ground. So we can use the following suvat equation:

$v=u+at$

where

v is the  final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

$a=g=9.8 m/s^2$ is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

$v=0+(9.8)(0.75)=7.35 m/s$

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