If the ball was thrown straight up at 16 ft/sec when it was 5 ft above the ground, how high did it go?

Question

A baseball is thrown in a parabolic arc.  It's position above the ground at a given point in time can be represented by the quadratic function       , where   ,  is -32 ft/sec/sec,   is initial velocity, and  is initial velocity, and is initial velocity, and  is its initial position above the ground.

  If the ball was thrown straight up at 16 ft/sec when it was 5 ft above the ground, how high did it go?  (Type the number, do not include units in your answer)

Answer 1

The time at which maximum height occurs can be found by setting the derivative equal to zero and solving for t. I will omit units because every value is given in ft and s, and the problem says to not include units in the answer.

p'(t) = g*t + v_o

0 = (-32)*t + 16

t = 0.5

Now to solve how high it went at this time, plug this time back into p(t):

p(0.5) = 1/2*(-32)*(0.5)² + (16)*(0.5) + (5)

p(0.5) = -4 + 8 + 5

p(0.5) = 9

This can be verified by plotting p(t). The max occurs at (0.5, 9)