Question

3, 2, 1 Bottle Rocket Science.. Show.yourwork for each pert... a previous ng worn at all times) from a high pressure launcher. The rocket was launched straight up to 220 ILL each group had launched a 2 L water bottle rocket (with safety goggles oei a height h. angle 8nax that it rose above the ground. The data is below. DATA Students were standing a horizontal distance d from the rocket and measured an Analysis Assume (incorrectly for simplification) that the rocket had a constant upward acceleration for the first 50 cm of ascent, at that height it lost all of its water-fuel. What max speed did the rocket achieve when it reached 50 cm above the launch pad? Vinst. at 5o What acceleration did the rocket have for the first 50 cm above the launch pad? Label each vector and include Draw a force diagram for the rocket when accelerating upward. its magnitude. Thrust

Answer 1

given, 2L water bottle launch

mr = 723.5 g

mwater = m

mbottle = 50 g

d = 22.97 m

thetamax =33.7 deg

Analysis

a. for constant upward acceelration for forst 50 m of ascent

let acceleration be a

now,

final total height h = d*tan(thetamax) = 15.319085703 m

maximu speed at h' = 50 cm = v

then

2*a*0.5 = v^2

also,

2*g*(h - 0.5) = v^2

hence

2*a*0.5 = 2*g(h - 0.5)

a = 290.75046150 m/s/s

v = 17.05140643 m/s

b. a = 290.75 m/s/s

c. force diagram:

thrust gravity weight

impulse : change in momentum = (mr + mbotlte)v = 13.189 Ns

average velocity of the water ejected = u

m *u = 13.189262873605

now,

from the initial conditions, m = 2 kg

hence

u = 6.5946314 m/s

rocket equation

using

change in momnetum of rocket = change on momentum of fuel

dP = dp

now, dP = mdV

dp = -u*dm

where m is mass of rocket at time t

hence

dP = dp

mdV = -udm

dm/m =- dV/u

integrating

uln(mf/mi) = v

where v is final speed of the rocket starting form rest

mf is final mas of rocket

mi is initial mass of rocket

u is exhaulst velocity of water