Question

A small rock is thrown vertically upward with a speed of 22.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock doesn’t hit the building on its way back down and lands on the street below. Ignore air resistance. (a) What is the speed of the rock just before it hits the street? (b) How much time elapses from when the rock is thrown until it hits the street?

Answer 1

Answer:

a) $v=32.74\frac{m}{s}$.

b) $t=\frac{v-v_0}{g} =5.58s$.

Explanation:

a) What is the speed of the rock just before it hits the street?

From Kinematics we have $v^2= v_0^2+2a(y_f-y_i)$.

If we take the top of the roof as the position $y_0=0m$, then

$y_f=-30m$ and we have (also, $a=g=-9.8\frac{m}{s^2}$) :

$v^2= v_0^2+ 2ay_f$ ⇒ $v= \sqrt{v_0^2+ 2ay_f}$

⇒ $v=32.74\frac{m}{s}$.

b) How much time elapses from when the rock is thrown until it hits the street?

From Kinematics we have $v(t)=v_0+at=v_0+gt$

When the rock touches the ground:

$v=-32.74\frac{m}{s}$

With a minus sign to indicate the vector velocity points down.

$t=\frac{v-v_0}{g} =5.58s$

(remember that $g=-9.8\frac{m}{s^2}$)