Question

An apple falls from a tree branch `3.5` `m` high to the ground. What is (a) the velocity of the apple just before it hits the ground and (b) the average force exerted on it by the ground if it decelerates to rest in `0.28` `s`?

Answer 1

Part A: `v=8.3` `ms^-1`
Part B: `F_(av)=6.4` `N`

Explanation:

The velocity of the apple will be given by:

`v^2=u^2+2as`, where the initial velocity `u` is `0` `ms^-1` and the acceleration is the acceleration due to gravity, `9.8` `ms^-1`.

`v^2=0^2+2xx9.8xx3.5=68.6`

`:.v=8.3` `ms^-1`

The momentum of the apple will be given by `p=mv=0.22xx8.3=1.8` `kgms^-1`

The impulse required to stop it will be given by `I=F_(av)t` and has the same magnitude as the momentum (in other words, the impulse acts to reduce the momentum from `1.8` `kgms^-1` to zero).

`I=F_(av)t`, rearranging:

`F_(av)=I/t=1.8/0.28=6.4` `N`