Question

You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 16.0 m above the ground, it is traveling at 22.0 m/s upward. Use the work-energy theorem to find the rock's speed just as it left the ground.

Answer 1

Answer:

The final speed of the stone as it lift the ground is 23.86 m/s.            

Explanation:

Given that,

Force acting on the rock, F = 3 N

Distance, d = 16 m

Initial speed of the stone, u = 22 m/s

We need to find the rock's speed just as it left the ground. It can be calculated using work energy theorem as :

$W=\Delta E\\\\W=\dfrac{1}{2}m(v^2-u^2)\\\\Fd=\dfrac{1}{2}m(v^2-u^2)\\\\v^2=\dfrac{2Fd}{m}+u^2\\\\v^2=\dfrac{2mgd}{m}+u^2\\\\v^2=2\times 9.8\times 16+(16)^2\\\\v=23.86\ m/s$

So, the final speed of the stone as it lift the ground is 23.86 m/s.