- #1

- 2

- 0

## 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

vf^2 = vi^2 + 2ad

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.