Question

A catapult launches a rocket at an angle of 47.5° above the horizontal with an initial speed of 109 m/s. The rocket engine immediately starts a burn, and for 2.49 s the rocket moves along its initial line of motion with an acceleration of 30 m/s2. Then its engine fails, and the rocket proceeds to move in free-fall.
(a) Find the maximum altitude reached by the rocket.



Your response differs from the correct answer by more than 10%. Double check your calculations. m
(b) Find its total time of flight.
s
(c) Find its horizontal range.
m

Answer 1

v(xo) = v(o)•cosα =109•cos47.5 = 73.6 m/s,
v(yo) = v(o)•sinα =109•sin47.5 = 80.4 m/s,
a(x) = a•cosα = 30•cos47.5 = 20.3 m/s^2,
a(y) = a•sinα =30• sin47.5 = 20.3 m/s^2.
There are three parts of rocket motion:
1. The upward motion with engine
v(x) = v(xo) + a(x)•t = 73.6+20.3•2.49=124.1 m/s.
v(y) = v(yo) + a(y)•t =80.4+22.1•2.49 = 135.4 m/s.
x1 = v(xo)•t + a(x)•t^2/2 = 73.6•2.49 + 20.3•2.49^2/2 = 246.2 m,
y1 = v(yo)•t + a(y)•t^2/2 = 80.4•2.49 + 22.1•2.49^2/2 = 268.7 m,
2. The upward motion
a(x) = 0, v(x) = 124.1 m/s.
v(y1) =v(y) – g•t1,
v(y1) = 0 => t1 =v(y)/g = 135.4/9.8 =13.8 s.
x2 =v(x)•t1 = 124.1•13.8 = 1714.6 m,
y2 = v(y) •t1 – g•(t1)^2/2 = 135.4•13.8 – 9.8•(13.8)^2/2 = 935.3 m.
3. The downward motion
h = y1+y2 =268.7+935.3 = 1204 m,
v(y2)^2 = 2•g•h = 2•9.8•1204 = 23598.4
v(y2) = sqrt23598.3 = 153.6 m/s,
v(y2) = g•t2, => t2 = v(y2)/g = 153.6/9.8 = 15.7 s,
v(x) = 124.1 m/s,
x3 = v(x) •t2 = 124.1•15.4 = 1945.3 m.
t(total) = 2.49 +t1+t2 = 2.49 +13.8 +15.7 = 32.5 s.
the range x = x1+x2+x3 =246.2 + 1714.6 + 1945.3 =3906.1 m.
ANS. (a) h = 1204 m, (b) t(total) = 32.5 s. (c) x = 3906.1 m.