Question

A 200 kg boulder is 1000m above the ground. What is the speed of the object right before it hits the ground?

Answer 1

Answer:

$140.07m/s$

Explanation:

we use the equation for the final speed:

$v_{f}^2=v_{0}^2+2gh$

where $v_{f}$ is the final speed, $v_{0}$ is the initial speed, $g$ is the gravitational acceleration ($g=9.81m/s^2$) and $h$ is the height.

In this case we don't have an initial velocity indicated so: $v_{0}=0$, and we are told that the boulder is at a height: $h=1,000m$.

We substitute this values into the equation:

$v_{f}^2=2gh\\v_{f}^2=2(9.81m/s^2)(1,000m)\\v_{f} ^2=19,620m^2/s^2\\\\v_{f}=\sqrt{19,620m^2/s^2} \\v_{f}=140.07m/s$

the speed of the object right before it hits the ground is $140.07m/s$