Question

4. A safe is hurled down from the top of a 1.3 x 102 m
building at a speed of 11.0 m/s. What is its velocity as
it hits the ground?​

Answer 1

The final speed of the safe is 51.7 m/s

Explanation:

The motion of the safe is a free fall motion, which is a motion at constant acceleration ($g=9.8 m/s^2$, towards the ground). Therefore, we can apply the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem we have (taking downward as positive direction):

u = 11.0 m/s

$a=g=9.8 m/s^2$

$s=1.3\cdot 10^2 m= 130 m$ is the vertical displacement (the height of the building)

Solving for v, we find the final velocity (and so, the final speed):

$v=\sqrt{u^2+2as}=\sqrt{11.0^2+2(9.8)(130)}=51.7 m/s$

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