a rocket has a mass of 70.kg and is initailly at rest on an inclined surface tha
ID: 1437434 • Letter: A
Question
a rocket has a mass of 70.kg and is initailly at rest on an inclined surface that rises 25 degress above the horizonatal. It is launched by applying a force of 2540 N parallel to the inclined surface, which accelerates it along a 220 m incline (starting from rest and at point A.) There is no friction between the rocket and the incline. The instant the rocket leaves the incline, the engines are turned off, it is subject only to gravity, and air resistance can be ignored
-What is the rocket acceleration?
-What is the rocket's velocity (magnitude) when it leaves the incline?
-Find the maximum height above the ground that the rocket reaches.
-Find the greatest horizontal range of the rocket beyond point A.
Explanation / Answer
a) vertical Acc. = a = 2540*cos25/70 = 32.9 m/s^2
b)
c) net acceleration (up along) = a =32.9 m/s^2
v^2 (launch) = 0 + 2 a s = 2 *32.9*220 = 14476
v (launch) = 120.3 m/s
height of launch = h(launch) = 220 * sin25 = 92.98 m
angle of lanuch = 25 deg
position and velocity components after time (t)
y - h = v sin 25 * t - 0.5 gt^2 ------ (1)
x = v cos 25 * t ---- (2)
Vx = dx/dt = v cos 25 ------(3)
Vy = dy/dt = v sin 25 - gt ------ (4)
at max height > Vy=0 = v sin 25 - gt
t = v sin 25/g >> put in (1)
h(max) = y = h + v sin 25 *[v sin 25/g] - 0.5g[v sin 25/g]^2
h(max) = h + v^2 sin^2 (25)/2g
h(max) = 92.98 + [14476*0.18/2*9.8] = 225.9 meter
d) Range
for time of flight (T) >>> putting y=0 >> or rocket hits the ground
(1) >>
0 - h = v sin 25 * T - 0.5 g T^2
gT^2 - 2v sin 25 * T - 2h = 0 ------ (5)
9.8 T^2 - (101.68) T - 185.96 =0
solving >. T = 11.96 s (leaving -ve time)
R = v cos 25 * T
R = 120.3 [cos 25] *11.96
R = 1303.98 meter
Range from point A (rest) = p + R
where p = 220 cos 25 = 199.4 m
R(t) = 1503.38 meter
Hope it is correct.
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