Question

A rock is thrown downward from the top of a 40.0-m-tall tower with an initial speed of 12 m/s. Assuming negligible air resistance, what is the speed of the rock just before hitting the ground?

(a) 28 m/s

(b) 30 m/s

(c) 56 m/s

(d) 784 m/s

(e) More information is needed.

Short answer

The speed of the rock just before hitting ground is $30\text{m}/\text{s}$ and the option (b) is correct.

Step-by-step solution

Identify given data

The height of tower is $h=40\text{m.}$

The initial speed of rock is $u=12\text{m}/\text{s}$.

Determine speed of rock just before hitting ground

The speed of the rock just before hitting the ground is found by using the third equation of motion with gravity.

The speed of the rock just before hitting ground is given by

$v^2=u^2+2gh$

Here, is the gravitational acceleration and its value is $9.80\text{m}/{\text{s}}^2$.

Substitute all the values in the above equation.

localid="1663740707419" $$\begin{gathered} v^2={\left(12\text{m}/\text{s}\right)}^2+2\left(9.80\text{m}/{\text{s}}^2\right)\left(40\text{m}\right) \\ v\approx 30\text{m}/\text{s} \end{gathered}$$

Therefore, the speed of the rock just before hitting ground is$30\text{m}/\text{s}$ and the option (b) is correct.