y = 5(10 + 4t - t^2)
max height achieved where t = -b/2a = 2
y(2) = -5(4) + 20(2) + 50 = 70
as above, when t=2
The rooftop is the ball's initial height. That is, when t=0. So, 50 ft.
The height of a ball thrown vertically upward from a rooftop is modelled by y=-5t^2+20t+50 , where h(t) is the ball's height above the ground , in meters , at time t seconds after the throw
a) Determine the max height of the ball
b) how long does it take for the ball to reach its max height
c) How high is the rooftop
2 answers
thank you so much
1 answer