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.
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.