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?

Short Answer

Expert verified

The acceleration of the rocket is 5.5 m/s2

Step by step solution

01

Draw a diagram to represent the problem:

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:
t0=0s,u=0m/s,y0=0m,a0=?t1=4s,a1=-g,v1=?,y1=?t2=6s,a2=-g,v2=0m/s,yf=0m


02

Calculating the acceleration of the rocket:

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

v1=u+a0t1

Substitute the known values into the above equation to solve for v1.
v1=0+a0(4)v1=4a0

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,

y1=y0+ut1+12a0t12y1=0+0(4)+12a0(4)2y1=8a0

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:

yf=y1+v1(t2)+12a1(t2)2

Substitute the known variables into the above equation
0=8a0+4a0(6)+12(-10)(6)2a0=5.5m/s2

Thus the acceleration of the rocket is 5.5 m/s2

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