Question

An object is dropped from a​ tower, 576 ft above the ground. The​ object's height above ground t seconds after the fall is ​s(t)=576−16t2. Determine the velocity and acceleration of the object the moment it reaches the ground.

Answer 1

Step-by-step explanation:

We are looking to solve for final velocity in this problem. We know the initial velocity is 0 since someone was holding the object before it was dropped, the displacement is -576 feet since the object had to fall that far below the point from which it was dropped (which is why it is negative), and the acceleration AT ANY POINT DURING THE FALL IS -32 ft/s/s. The equation we use to find final velocity is

v² = v₀² + 2aΔx and filling in:

v² = 0² + 2(-32)(-576) so

`v=√(2(-32)(-576))` so

v = 192 ft/s