Rocket Acceleration: Kinematics Problem Solved

[mjolnir80]

Homework Statement

a rocket is launched straight up with a constant acceleration. four seconds after takeoff , a bolt falls off the side of the rocket. the bolt hits the ground 6.0 seconds later. what was the rockets acceleration?

Homework Equations

any kinematics equations

The Attempt at a Solution

we know that the initial velocity of the bolt equals the final velocity of the rocket also
thats as far as i can get

thanks in advanced for the help

 

[Monocles]

Once the bolt leaves the rocket you will have the bolt's velocity and acceleration due to gravity, and how long it will take to hit the ground. You can use this information to find how high the bolt was when it left the rocket. Once you have that information, you will be able to find the acceleration the rocket must have had to get to that height in four seconds.

 

[mjolnir80]

Once the bolt leaves the rocket you will have the bolt's velocity and acceleration due to gravity, and how long it will take to hit the ground. You can use this information to find how high the bolt was when it left the rocket. Once you have that information, you will be able to find the acceleration the rocket must have had to get to that height in four seconds.

actually you don't have the bolts velocity

 

[Galileo's Ghost]

Keep in mind the delta y for the rocket in the first 4 seconds will have the same magnitude but opposite direction as the delta y for the bolt in the final 6 seconds. So (delta y)rocket = -(delta y)bolt. Also the vf for the rocket (and consequently the vi of the bolt) can be solved in terms of a using the kinematic vf = vi + at ...that value for vf of the rocket can be used in the expression for (delta y)rocket on the left and can be used as vi of the bolt in an expression for (delta y) on the right.

 

[Display MOB]

Isn't acceleration always -9.8m/s^2 (If your positive direction is up)?

Also what does

"a bolt falls off the side of the rocket. the bolt gits the ground 6.0 seconds later."

have to do with the rockets acceleration?

 

[Galileo's Ghost]

Isn't acceleration always -9.8m/s^2 (If your positive direction is up)?

Also what does

"a bolt falls off the side of the rocket. the bolt gits the ground 6.0 seconds later."

have to do with the rockets acceleration?

Acceleration is 9.80 m/s^2 downward for objects that are in free fall. The rocket is not in free fall. It is burning fuel to generate a thrust accelerating the rocket upwards. The bolt IS in free fall over the last 6 seconds. During the first 4 seconds, however, the bolt was experiencing the exact same acceleration as the rocket.

 

[Display MOB]

"The rocket is not in free fall. It is burning fuel to generate a thrust accelerating the rocket upwards."

Thanks, I'm new to this.

 

[mjolnir80]

Keep in mind the delta y for the rocket in the first 4 seconds will have the same magnitude but opposite direction as the delta y for the bolt in the final 6 seconds. So (delta y)rocket = -(delta y)bolt. Also the vf for the rocket (and consequently the vi of the bolt) can be solved in terms of a using the kinematic vf = vi + at ...that value for vf of the rocket can be used in the expression for (delta y)rocket on the left and can be used as vi of the bolt in an expression for (delta y) on the right.

do we assume that Vi of the rocket is zero?

 

[mridkash]

velocity of the bolt going up with rocket after t sec, $$v_{up} = at_{up}$$ (up positive)
initial velocity is zero here as the rocket starts from rest.

Now this velocity becomes the initial velocity of the bolt when it starts to fall down.

When the bolt is falling, it is acted upon by -g acceleration, so the distance traveled by the falling bolt is given by $$s_{down} = (at_{up})t_{down} - \frac{1}{2}gt_{down}^{2}$$ where $$t_{down}$$ is the time taken by the bolt to hit ground.

But we know already that, distance the bolt travels up and down are same only the times differ $$s_{down} = s_{up} = \frac{1}{2}at_{up}^2$$

Putting the value of distance in former equation we get,

$$\frac{1}{2}at_{up}^2 = (at_{up})t_{down} - \frac{1}{2}gt_{down}^{2}$$
the times are given and g you know.