Question

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.10s later. What was the rocket's acceleration?

Answer 1

1.3m/s²

Step-by-step explanation:

Given values in the first part: acceleration a, time t₁:

1) velocity v₀
`= at_1`

2) height h₀
`=(1)/(2)at_1^2`

Given values in the second part: acceleration -g, time t₂:

3) height h
`= -(1)/(2)gt_2^2+v_0t_2+h_0`

Combining equations 1,2,3 and setting h to zero:

`0=-(1)/(2)gt_2^2+(at_1)t_2+(1)/(2)at_1^2\\ 0=a(t_1t_2+(1)/(2)t_1^2)-(1)/(2)gt_2^2`

Solve for a with t₁ = 4s and t₂=2.1s:

`a=(1)/(2)gt_2^2((1)/(t_1t_2+(1)/(2)t_1^2))`