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

Answer:

1.3m/s²

Explanation:

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

1) velocity v₀ $= at_1$

2) height h₀ $=\frac{1}{2}at_1^2$

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

3) height h $= -\frac{1}{2}gt_2^2+v_0t_2+h_0$

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

$0=-\frac{1}{2}gt_2^2+(at_1)t_2+\frac{1}{2}at_1^2\\ 0=a(t_1t_2+\frac{1}{2}t_1^2)-\frac{1}{2}gt_2^2$

Solve for a with t₁ = 4s and t₂=2.1s:

$a=\frac{1}{2}gt_2^2(\frac{1}{t_1t_2+\frac{1}{2}t_1^2})$