A ball is thrown in the air (assuming from ground level). The path of the ball follows the position function: y = 100t-16t^2. What is the terminal velocity of the ball?

Question

anonymous · Answer

v=y' =100-32t ==> v(fianal means y =0) ==> 100t -16t^2=0 ==>4t(25-4t)=0 ==> t= 0 or t=25/4  ===> v(final at t=25/4) = 100-32(25/4) = -100 ( the minus sign means in the opposite dirction , i.e. falling down)

anonymous · Answer

The answer is correct ^^ However, I don't understand why you would plug (25/4) into t.

anonymous · Answer

Because (25/4) is the time when the terminal velocity occurs, not the terminal velocity itself.

anonymous · Answer

well , t=25/4 it's her the time at which the terminal velocity occured.

anonymous · Answer

the terminal velocity means the position = 0 ==> solve for  posotion function y =0 to find the time at which the terminal velocity occurs, then find the 1st dervative of y which will be the velocity relation, plug t= 25/4 in it to find the terminal velocity at the time it happened. so we got two works, the 1st one is to find the time then to plug into the derivative.