How do I calculate the height this water rocket will rise?

• Sugarcrashed
In summary: I get 3.4 m.In summary, a 15.0 g plastic water rocket containing 40.0 g of water and compressed air is fired on a low friction horizontal wire. The compressed air forces the water out of the rocket in 0.600 s at an average exit velocity of 4.05 m/s [E]. Using formulas for acceleration, net force, distance, velocity, and momentum, the rocket can be calculated to have an initial velocity of 10.8 m/s [W] after the 0.6 seconds. However, the given answer of 6 m for the height the rocket would have risen if directed upwards seems incorrect. The correct answer is likely closer to 3.4
Sugarcrashed

Homework Statement

A 15.0 g plastic water rocket containing 40.0 g of water and compressed air is strung on a very low friction horizontal wire. When fired, compressed air forces the water from the rocket in 0.600 s at an average exit velocity of 4.05 m/s [E].

Based on the information in the above narrative, if the rocket had been directed upwards, approximately how high would it have risen?

The answer is supposed to be 6 m high, but I cannot figure out how to get it.

Homework Equations

V = d/t
a = v/t
a = Fnet/m
g = Fg/m
d = vi*t + 1/2at^2
d = vf*t - 1/2at^2
d = ((vf + vi)/2)t
Ek = 1/2mv^2
Ep = mgh
W = delta E

The Attempt at a Solution

First I calculated the acceleration of the water with the formula a = v/t, and it gave me 6.75 m/s^2 [E].
4.05 m/s [E] / 0.600 = 6.75 m/s^2 [E]

Then I changed all the masses into kg.
15.0 g = 0.015 kg
40.0 g = 0.04 kg

Then I calculated the net force of the water with the formula F = ma, and it gave me 0.27 N [E].
0.04 kg * 6.75 m/s^2 [E] = 0.27 N [E]

Newtons law states that for every action, there is an equal and opposite reaction, meaning that there would also a be a force of 0.27 N to the west. (The force that would make the rocket propel.)

I then calculated the acceleration of the rocket using the formula a = Fnet/m, giving me 18 m/s^2 [W].
0.27 [W] / 0.015 kg = 18 m/s^2 [W]

If the rocket is directed upwards, it is now fighting against gravity, so I subtracted 9.81 m/s^2 [D] from my 18 m/s^2 [Up], giving me 8.19 m/s^2 [Up].
18 m/s^2 [Up] - 9.81 m/s^2 [D] = 8.19 m/s^2 [Up]

Using the formula d = Vi * t + 1/2at^2, I calculated the distance between the ground and the point when the rocket finished propelling all of its water out, giving me 1.47 m.
0 m/s [Up] * 0.600 s + 1/2( 8.19 m/s^2 [Up] * 0.600 s^2 ) = 1.47 m

Using the formula Vf = Vi + at, I calculated the velocity when the rocket finished propelling all of its water out, giving me 4.92 m/s [Up].
0 m/s [Up] + 8.19 m/s^2 [Up] * 0.600 s = 4.92 m/s [Up]

I then used the formula d = Vf^2 - ( Vi^2 / 2a ) to calculate the distance the rocket when up after propelling all of its water out.
0 m/s [Up] ^2 - ( 4.92 m/s [Up]^2 / (2 * 9.81 m/s^2 [D] )) = 1.23 m

I then added the two distances together, giving me 2.7 m. Apparently, this is incorrect.
1.47 m + 1.23 m = 2.7 m.

Welcome to PF

It may be simpler if you find the initial velocity (after the 0.6 seconds) from momentums instead of forces. What is the momentum of the system after the 0.6 seconds? What is the momentum of just the rocket?

By using the formula P = mv, I am able to find the momentum of the water leaving rocket.
0.04 kg * 4.05 m/s [E] = 0.162 kg*m/s [E]

I am assuming that momentum is similar to force and would give the the rocket a momentum of 0.162 kg*m/s [W].
I could then find the velocity of the rocket with the formula v = P/m.
0.162 kg*m/s [W] / 0.015 kg = 10.8 m/s [W].

I used the formula d = Vf^2 - ( Vi^2 / 2a ) to calculate the distance after all the water was emptied out.
0 m/s [Up] ^2 - ( 10.8 m/s [Up]^2 / (2 * 9.81 m/s^2 [D] )) = 5.94 m

I added the two distances together, and again I did not get 6 m. At least it is closer than my last attempt.
5.94 m + 1.47 m = 7.41 m

The 5.94 m by itself is the correct answer, but shouldn't that be the distance it moved after the 0.6 seconds?

Sugarcrashed said:
First I calculated the acceleration of the water with the formula a = v/t, and it gave me 6.75 m/s^2
A classic case of plugging in numbers just because they're the right units for an equation, without regard to what the equation actually requires the numbers to mean.
Does the water accelerate from 0 to 4.05 m/s gradually over a period of 0.6 seconds?
Sugarcrashed said:
0.04 kg * 4.05 m/s [E] = 0.162 kg*m/s [E]
Well, not really. The 'exit speed' of the water should be relative to the bottle, but I suspect the author of the question intends to be kind to you here and wants you to use this as speed relative to the ground, as you have done.
Sugarcrashed said:
I used the formula d = Vf^2 - ( Vi^2 / 2a ) to calculate the distance after all the water was emptied out.
But what is the speed after all water has been emptied out? Don't forget that gravity has been acting on the bottle all this time.

By the way, even allowing the given water exit speed as being relative to the ground, I disagree with the given answer. It's way too high.

Last edited:

1. How do I calculate the launch angle for my water rocket?

The launch angle for a water rocket can be calculated using the formula tanθ = (h/d), where θ is the launch angle, h is the height the rocket will reach, and d is the distance from the launch point to the point where the rocket will land. Rearranging the formula to solve for θ, the launch angle can be calculated as θ = arctan(h/d).

2. What is the equation for calculating the maximum height of a water rocket?

The maximum height of a water rocket can be calculated using the formula h = (v^2*sin^2θ)/2g, where v is the initial velocity of the rocket, θ is the launch angle, and g is the acceleration due to gravity (9.8 m/s^2). This formula assumes that there is no air resistance acting on the rocket.

3. How do I determine the initial velocity of my water rocket?

The initial velocity of a water rocket can be calculated using the formula v = √(2gh), where v is the initial velocity, g is the acceleration due to gravity, and h is the maximum height the rocket will reach. This formula assumes that there is no air resistance acting on the rocket.

4. Can I use the same equation to calculate the height of my water rocket if there is air resistance?

No, the equations for calculating the height of a water rocket assume there is no air resistance. If air resistance is present, the equations become more complex and may require advanced mathematical techniques to solve.

5. How can I improve the accuracy of my calculated water rocket height?

To improve the accuracy of your calculated water rocket height, you can conduct multiple launches and take an average of the results. You can also make sure to carefully measure the launch angle and initial velocity of the rocket. Additionally, you can consider factors such as air resistance and wind when conducting your calculations.

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