A player bumps a volleyball with an initial vertical velocity of 20 ft/s. The height of the ball can be model by the function h(t) = -16t2 + 20t + 4, where h is the height in feet and t is the time in seconds.

Question

a. What is the maximum height of the ball? ( use your graphing calculator or find the vertex)

d. Suppose the volleyball were hit under the same conditions, but with an initial velocity of 32 ft/s. How much higher would the ball go?

Answer 1

your equation should have been
h(t) = -16t^2 + 20t + 4

a) It is your graphing calculator, I don't know what you have nor do I know how to use yours.
Follow the steps outlined in the manual to answer a)

OR

for f(x) = ax^2 + bx + c
the x of the vertex is -b/(2a)
so for yours, the t of the vertex is -20/(-32) = .625
h(.625) = -16(.625^2) + 20(.625) + 4 = 10.25
so the vertex is (.625, 10.25)
then the maximum height is 10.25 ft.

OR

h '(t) = -32t + 20 = 0 for a max/min of h(t)
32t = 20
t = .625 as above, find h(.625)

OR
complete the square, a method you might have learned in your course.

b) change 20t to 32t in your equation and repeat a)

Answer 2

What I’m the world. help with the question.