Since velocity is the derivative of position,
v(t) = -32t + 160
now just solve
-32t+160 = 32
-32t+128 = 0
t = 4
The height in feet above the ground of a ball thrown upwards from the top of a building is given by s=-16t^2 + 160t + 200, where t is the time in seconds.
If the maximum height is 600 feet, what is v^-1(32)?
The answer is supposed to be 4 seconds, but I don't understand how. please explain
2 answers
Thank you very much!