Question

A cherry falls from a tree branch that is 9 feet above the ground, how far above the ground is the cherry after 0.2 seconds?

a cherry falls from a tree branch that is 9 feet above the ground, how far above the ground is the cherry after 0.2 seconds?

Answer 1

8.36 feet

Explanation:

SUVAT equations

If the cherry is undergoing free fall with negligible air resistance we can assume that the only acceleration is acceleration due to gravity which is `9.81 ms^{-2}`

Assuming the cherry's height is the same as that of the tree branch it is 9 ft above the ground, or `2.74 m` in SI units.

`s=ut + 1/2at^2` can be used to find the displacement for a body with uniform acceleration.

Time `t=0.2s`
Acceleration `a=9.81 ms^{-2}`
Initial velocity `u=0`, since we assume the cherry isn't moving until it starts falling.
This gives us all the values we need to find the displacement `s`

`s=1/2 times (9.81) times (0.2)^2 = 0.1962 m` which is equal to `0.64` feet

Now, notice that this is the displacement from the branch, not distance from the ground. To find that we subtract the distance fallen from the height of the branch.

`9-0.64=8.36` feet