How far is the rock from the ground one second after it is dropped?

Question

A rock is dropped from a height of 50 feet. Its speed in feet per second at time t after it is dropped is given by
s(t)=−32t

where time t is measured in seconds. How far is the rock from the ground one second after it is dropped?

Answer 1

_{Displacement is the integral of velocity. Therefore, we integrate the given velocity function with respect to t.}

 

∫s(t) = -16t² + d_i

 

d_i is the initial distance from zero, which is given as 50 feet. So we have:

 

d(t) = -16t² + 50

 

So at time t=1, we have:

 

d(1) = -16(1) + 50

d(1) = -16 + 50 = 36 feet

 

Our final answer is 36 feet.