Question

A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What is the rock's velocity as it hits the bottom of the hole?b.) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Answer 1

(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

$K=\frac{1}{2}mv^2$

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

$v^2 = u^2 + 2gd$

where

u = 21 m/s is the initial speed of the rock as it enters the hole

g = 9.8 m/s^2 is the acceleration due to gravity

d = 10 m is the depth of the hole

Substituting,

$v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s$

(b) 4.72 s

The vertical position of the rock at time t is given by

$y(t) = v_y t - \frac{1}{2}gt^2$

where

$v_y = 21 m/s$ is the initial vertical velocity

Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:

$-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0$

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution