Question

A toy rocket is launched straight up into the air with an initial velocity of 60 ft/s from a table 3 ft above the ground. If acceleration due to gravity is –16 ft/s2, approximately how many seconds after the launch will the toy rocket reach the ground?

h(t) = at2 + vt + h0
0.05 s
2.03 s
3.80 s
3.70 s

Answer 1

h(t) = -16t^2 + 60t + 3. Set h(t)=0:

t = [ -60 ± sqrt(60^2 - 4(-16)(3)) ] / (2(-16))
= [ -60 ± sqrt(3792) ] / -32
= (60 ∓ sqrt(3792)) / 32.

The positive root ≈ (60 + 61.592) / 32 ≈ 3.80 s (the other root ≈ -0.05 s is before launch).

Answer: 3.80 s.