91Ó°ÊÓ

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