An object is dropped from a tower, 1296ft above the ground. The objects height is above ground t seconds after the fall is s(t)=1296-16t^2

Question

^{Determine the velocity and acceleration of the object the moment it reaches the ground} 

Answer 1

Since this is listed as a calculus problem, let us work with derivatives.

 

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

 

a(t) = dv/dt = -32

 

Set s(t) = 0, solve for t, and evaluate v(t) and a(t) at this value of t.

 

1296 - 16t² = 0

 

t² = 1296/16 = 81

 

t = 9 seconds

 

Evaluate v(9) and a(9).  Just plug in the values and finish from here.