SOLUTION: This is from a test question which I struggled with. Use the projectile motion equation: h= -16t^2+V0t+h0 A baseball is hit from a rooftop 64 feet above the ground with an in

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Question 1149884: This is from a test question which I struggled with.
Use the projectile motion equation: h= -16t^2+V0t+h0
A baseball is hit from a rooftop 64 feet above the ground with an initial upward velocity of 160 feet per second.
A. How long after the ball is hit will it be 128 feet above the ground?
B. How long after the ball is hit will it hit the ground?
C. What is the height of the ball 2 seconds after it was hit?

Answer by ikleyn(52855)   (Show Source): You can put this solution on YOUR website!
.

(A)  To answer (A), you need to substitute  h= 128 into the formula for the current height; 

     v0= 160 (initial upward velocity) and h0= 64 (initial height).


     Then you will get the equation


         -16t^2 + 160t + 64 = 128.      (A)


     I will left it to you to solve it on your own.



(B)  To anser (B), you need to solve the equation


                  -16t^2 + 160t + 64 = 0.    (B)


     In this case, the current height is 0 (zero), since the ball hits the ground (which is "level zero").



(C)  To answer this question, you need substitute t= 2 (2 seconds) into the given function


         h = -16*2^2 + 160*2 + 128.

What's all.



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