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
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.
To solve this problem, we can break it down into different stages: the projectile motion while the rocket is moving upward, the projectile motion while the rocket is moving downward, and the free-fall motion after the rocket engine fails.
(a) Finding the maximum altitude reached by the rocket:
First, we need to find the time it takes for the rocket to reach its maximum altitude. To do this, we can use the formula:
t = (Vf - Vo) / a
Where:
Vf = final velocity = 0 m/s (when the rocket is at its maximum altitude)
Vo = initial velocity = 109 m/s
a = acceleration = -9.8 m/s^2 (since it's acting opposite to the motion)
Plugging in the values, we get:
t = (0 - 109) / -9.8
t ≈ 11.12 s
Next, we can calculate the maximum altitude reached by the rocket using the formula:
h = Vo * t + (1/2) * a * t^2
Where:
h = maximum altitude
Vo = initial velocity = 109 m/s
t = time taken to reach maximum altitude = 11.12 s
a = acceleration = -9.8 m/s^2
Plugging in the values, we get:
h = 109 * 11.12 + (1/2) * (-9.8) * (11.12)^2
h ≈ 662.49 m
Therefore, the maximum altitude reached by the rocket is approximately 662.49 meters.
(b) Finding the total time of flight:
The total time of flight can be calculated by adding the time it took for the rocket to reach its maximum altitude (11.12 s) and the time it takes for the rocket to fall back to the ground.
To find the time it takes to fall back to the ground, we can use the formula:
t = sqrt(2h / g)
Where:
h = maximum altitude reached = 662.49 m
g = acceleration due to gravity = 9.8 m/s^2
Plugging in the values, we get:
t = sqrt(2 * 662.49 / 9.8)
t ≈ 11.12 s
Therefore, the total time of flight is approximately 11.12 + 11.12 = 22.24 seconds.
(c) Finding the horizontal range:
The horizontal range can be found using the formula:
R = Vx * t
Where:
R = horizontal range
Vx = horizontal component of initial velocity
To find Vx, we can use the equation:
Vx = Vo * cos(theta)
Where:
Vo = initial velocity = 109 m/s
theta = launch angle = 47.5°
Plugging in the values, we get:
Vx = 109 * cos(47.5°)
Vx ≈ 74.31 m/s
Now we can find the horizontal range using the formula:
R = Vx * t
R = 74.31 * 22.24
R ≈ 1652.65 m
Therefore, the horizontal range of the rocket is approximately 1652.65 meters.