Question

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)equals400 minus 16 t squared. Determine the velocity and acceleration of the object the moment it reaches the ground. The velocity of the object the moment it reaches the ground is nothing ​ft/s.

Answer 1

Answer: v= 160ft/s

a=32ft/s^2 constant

Explanation:

s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.

0=400-16t^2

400=16t^2

25=t^2

t=5s

ds/dt=v(t)=0-32t

dv/dt=a(t)=-32 constant(gravity)

v(t)=-32(5s)= -160ft/s negative sign is only showing direction