The height in feet of a projectile at time t seconds is h(t)= -16^2+480t+5 . Find the acceleration of the projectile when t = 3 seconds.

Answer 1

acceleration is a constant -32 feet per second per second, due to gravity

h(t)=-gt^2/2+vot+ho

take the 2nd derivative of h(t)=-16t^2+ 480t+5

h'(t) =v(t)=-32t +480

a(t)=v'(t)=h"(t) =g = -32 ft/sec^2

h"(3)=-32

at any time t, h"(t)=-32