physics
posted by tina .
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 freefall.
(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

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.