A ball is thrown upward into the air with an initial velocity of 64 feet per sec

to calculate height after 1.5 seconds, put in t =1.5 in equation h(t) = 64t - 16t2  and you will get your h(1.5)
to get maximum height, just differentiate h(t) wrt t and equate it to zero
ie by doing so, you would get at t=2 , the height is maximum.
after 4 seconds, the ball returns to the ground., you can do it by either doubling the time it takes the ball to get maximum height or by solving the quadratic h(t) = 64t - 16t2 =0
giving you two values of t, one being t=0 (obvious since the ball's height at t=o was 0 and t=4, the time that we found out by doubling the time taken to reach maximum height

Thank you, I tried factoring the quadratic, and it also helped: h(t) = 16t(4 - t2)... This is factored form, so I graphed it by finding the x-intercepts, vertex, and y-intercept.

Happy now

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marking topic as solved