Question

The function H(t) = -16t2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.

Answer 1

Part A:
The height; \(H\), of the object, at a time; \(t\), is
\[H(t) = -16t^2 + vt + s \]
when the object hits the ground, \(H(t=t_f) = 0\)

Answer 2

so solve: \[-16t_f^2+vt_f+s=0\] for \(t_f\),

with \(s=100\), \(v=60\)

Answer 3

time to max height ( above s not given)
-16t^2=100t
-16t=100
t=6.25 sec.