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 6.0s later. What was the rocket's acceleration?

In the given below diagram, the rocket goes upwards and the bolt falls back to the ground at time t₁.

The motion is divided into three phases:

1. A constant acceleration of the rocket upwards till time t₁.

2. The bolt is decelerated due to gravity up to the maximum height.

3. A uniform accelerated motion in which the bolt falls back to the ground at t₂.

Given Data:
$$\begin{gathered} t_0=0s,u=0m/s,y_0=0m,a_0=? \\ {t}_1=4s,a_1=-g,v_1=?,y_1=? \\ {t}_2=6s,a_2=-g,v_2=0m/s,y_f=0m \end{gathered}$$

The kinematic equations of motion to calculate the velocity at which the bolt exits are given by :

$v_1=u+a_0{t}_1$

Substitute the known values into the above equation to solve for v₁.
$$\begin{gathered} v_1=0+a_0\left(4\right) \\ {v}_1=4a_0 \end{gathered}$$

Thus, the velocity of the rocket at this point is 4a₀

The distance covered by the rocket at this point is y₁

Using the equation of motion,

$$\begin{gathered} y_1=y_0+u{t}_1+\frac{1}{2}{a}_0{t_1}^2 \\ {y}_1=0+0\left(4\right)+\frac{1}{2}{a}_0{\left(4\right)}^2 \\ {y}_1=8a_0 \end{gathered}$$

Thus, the distance covered by the rocket when the bolt falls down to the ground is 8a₀

The kinematic equations of motion to calculate the velocity at maximum altitude are given by:

$y_f=y_1+v_1\left(t_2\right)+\frac{1}{2}{a}_1{\left(t_2\right)}^2$

Substitute the known variables into the above equation
$$\begin{gathered} 0=8a_0+4a_0\left(6\right)+\frac{1}{2}(-10){\left(6\right)}^2 \\ {a}_0=5.5m/s^2 \end{gathered}$$

Thus the acceleration of the rocket is 5.5 m/s²