Question

A rock is thrown downward from a height of 207 m with an initial velocity of 5.37 m/s. Approximately how fast will it it be moving 3.03 seconds after it was thrown? (Assume no air resistance.)

Answer 1

List the known information:

$x_0=207 \text{ m}\\v_0=-5.37 \text{ m/s}\\t=3.03 \text{ s}\\a=-g=-9.8 \text{ m/s}^2$

Use the kinematic equation $v=v_0+at$.

Plug in the given values:

$v=-5.37 \text{ m/s}+(-9.8 \text{ m/s}^2)(3.03 \text{ s})=-35.064 \text{ m/s}$

This would be 35.064 m/s downward, or 35 m/s downward with significant figures taken into account.