An object is dropped from a tower 400 ft above the ground Th

An object is dropped from a tower, 400 ft above the ground. The object\'s height above ground t seconds after the fall is s(t) = 400 - 16t2. Determine the velocity and acceleration of the object the moment it reaches the ground.

Solution

velocity= d(s(t))/dt = -32t acceleration = d(velocity)/dt = -32ft/s^2 time taken to reach ground => s(t)=0 => 400= 16 t^2 => t=5s so velocity= -32*5=-160 ft/s and acceleration= -32ft/s^2