Question

Ryan throws a tennis ball straight up into the air. The ball reaches its maximum height at 2 seconds. The approximate height of the ball x seconds after being thrown is shown in the table.


Which equation models the motion of the ball?

Answer 1

well, you know that

y = vt - 4.9t^2

max height is at t = v/9.8

Answer 2

let the initial velocity be k m/s
so
height = -4.9t^2 + kt
d(height)/dt = -9.8t + k
when t = 2, d(height)/dt = 0
2 = -9.8(2) + k
k = 19.6

so your equation was
height = -4.9t^2 + 19.6t
the maximum height is
= -4.9(4) + 19.6(2) m
= appr 19.6 m

I used the more traditional variable t for time instead of the suggested x, no big deal.