A rocket is launched at an angle of 42◦ above
the horizontal with an initial speed of 76 m/s.
It moves for 9 s along its initial line of motion
with an acceleration of 29 m/s^2. At this time
its engines fail and the rocket proceeds to
move as a projectile.
The acceleration of gravity is 9.8 m/s^2 .
a.) Find the maximum altitude reached by the rocket.
b.) What is its total time of flight?
c.) What is its horizontal range?
The answers are:
a.) 3837.91 m
b.) 59.9965 s
c.) 14152.7 m
I'm not sure how to get these because of the increasing acceleration; I know that for part a slope=0, part b height=0, and I guess part c involves using the answer for part b.
The website for these examples is webphysics.iupui.edu/152/Hw/HW04.pdf
questions #6-8
To solve these problems, we need to break down the motion of the rocket into two parts: the time when the engines are functioning (with constant acceleration) and the time when the rocket moves as a projectile (with constant velocity). Let's go step by step:
a.) To find the maximum altitude reached by the rocket, we need to determine the rocket's vertical displacement at the end of the engine functioning phase.
1. First, we need to find the time it takes for the rocket to reach the maximum altitude. The initial velocity in the vertical direction is given by:
Vy0 = V0 * sin(theta) = 76 m/s * sin(42°) ≈ 49.29 m/s
We can use the equation:
Vy = Vy0 + a * t
Rearranging the equation to solve for time (t):
t = (Vy - Vy0) / a
t = (0 - 49.29 m/s) / (-9.8 m/s^2) ≈ 5.03 s
2. Now that we have the time it takes for the rocket to reach the maximum altitude, we can determine the vertical displacement using the equation for displacement:
y = Vy0 * t + (1/2) * a * t^2
Substituting the given values:
y = 49.29 m/s * 5.03 s + (1/2) * (-9.8 m/s^2) * (5.03 s)^2 ≈ 3837.91 m
Therefore, the maximum altitude reached by the rocket is approximately 3837.91 m.
b.) To find the total time of flight, we need to calculate the time it takes for the rocket to reach the ground after the engine failure.
1. The total time of flight can be found by summing the time spent in the initial motion phase (9 s) with the time spent as a projectile after the engines fail.
2. The time spent as a projectile can be determined by using the vertical displacement formula:
y = Vy0 * t + (1/2) * g * t^2
Since the rocket will land at the same vertical position as it started, the vertical displacement (y) is zero. Rearranging the equation, we get:
(1/2) * g * t^2 = -Vy0 * t
Simplifying further:
(1/2) * g * t = -Vy0
t = (-2 * Vy0) / g
t = (-2 * 49.29 m/s) / (-9.8 m/s^2) ≈ 10.04 s
3. Adding the time spent in the initial motion phase, the total time of flight is:
Total time = time with engine functioning + time as a projectile
= 9 s + 10.04 s ≈ 19.04 s
Therefore, the total time of flight is approximately 19.04 s.
c.) To find the horizontal range, we can use the formula:
Range = Vx * t
Since there is no horizontal acceleration, the horizontal velocity (Vx) remains constant throughout the motion. The initial horizontal velocity can be calculated using:
Vx0 = V0 * cos(theta) = 76 m/s * cos(42°) ≈ 57.85 m/s
Substituting the given values:
Range = Vx0 * (time with engine functioning + time as a projectile)
= 57.85 m/s * (9 s + 10.04 s) ≈ 14152.7 m
Therefore, the horizontal range is approximately 14152.7 m.
I hope this explanation helps you understand how to solve these problems!