Question

A ball is launched from a platform. The equation for the ball’s height h at time t seconds after launch is h(t)=-4.9t^2+16.17t+71.05 where h is in meters. What is the initial height of the ball before launch?

Answer 1

71.055 [m].

Explanation:

1) according to the condition t=0 (initiat height means elapsed time is zero), then

2) h(0)=4.9*0+16.17*0+71.05; ⇔ h(0)=71.05 [m].