An object is dropped from a tower, 576576 ft above the ground. The object's height above ground t seconds after the fall is s (t) equals=576 minus 16 t squared576-16t2. Determine the velocity and acceleration of the object the moment it reaches the ground.

Answer: 192 ft/s

Explanation:

The initial height of the object is:

576ft above the ground.

The position equation is:

p (t) = - 16*t^2 + 576

in the position equation, we only can see the therm of the initial height and the term of the acceleration (that is equal to the gravitational acceleration g = 32 ft/s^2 over 2)

So we have no initial velocity, this means that at the beginning we only have potential energy:

U = m*g*h

where m is the mass of the object, g = 32m/s^2 and h = 576 ft.

Now, as the object starts to fall down, the potential energy is transformed into kinetic energy, and when the object is about to hit the ground, all the potential energy has become kinetic energy.

The kinetic energy equation is:

K = (m/2) * v^2

where v is the velocity of the object, then the maximum kinetic energy (when the object reaches the ground) is equal to the initial potential energy:

m*g*h = (m/2) * v^2

now we can solve this for v.

v = √ (2*g*h) = √ (2*32ft/s^2*576ft) = 192 ft/s