Question

There was this apple that grew to an incrediable mass of1.67 kg. You are examining said apple, when you accidentally drop it. You manage to react quickly and catch it just before it hits the ground. The apple has achieved a speed of 5.61 m/s by this time. What is the K.E. of the apple at this time? How far were you holding the apple from the ground when you lost your grip?

Answer 1

Given: $m = 1.67~\text{kg};~v = 5.61~\frac{\text{m}}{\text{s}}.$

Kinetic energy at the moment when its speed is equal to v, is:

$K = \frac{1}{2}mv^2 = \frac{1}{2}~*~1.67~\text{kg}~*~(5.61\frac{\text{m}}{\text{s}})^2 \approx 26.28~\text{J}.$

To answer the question about the height of the drop, let's refer to the basic kinematic equations for the coordinate and for the velocity:

$\begin{cases} h = h_0 + v_0t + \frac{1}{2}gt^2; \\ v = v_0 + gt, \end{cases}$

where $g \approx 9.81~\frac{ \text{m} }{ \text{s}^2 }$ is the acceleration of gravity, and t is time.

We will assume that the apple was dropped from rest (thus initial velocity $v_0 = 0$) and we will take the initial position of the apple as zero as well ($h_0 = 0$).

Solving these equations together now and excluding the time t, we get:

$t = \frac{v}{g}; \\ h = \frac{v^2}{2g} = \frac{(5.61~\frac{\text{m}}{\text{s}})^2}{2~*~9.81~\frac{\text{m}}{\text{s}^2}} \approx 1.60~\text{m}.$

Pay attention that the answer for the height does not depend on the apple's mass, as this will be the same for any object in the Earth's gravity field. Also, we did neglect the air resistance.