A rocket is launched at an angle of 56.0° above the horizontal with an initial speed of 103 m/s. It moves for 3.00 s along its initial line of motion with an acceleration of 31.0 m/s2. At this time its engines fail and the rocket proceeds to move as a free body.
(a) Find the maximum altitude reached by the rocket.
m
(b) Find its total time of flight.
s
(c) Find its horizontal range.
m
On an extra note. I've done a lot of the practice problems and gotten those wrong... I don't know why I keep getting this wrong... But my answers for part a have been fairly close it seems like I keep getting my y-speed when the engine fails wrong by about 2.something and I don't know why but so far time has been around 36-37 seconds
To solve this problem, we can break it down into different parts.
(a) Find the maximum altitude reached by the rocket:
To find the maximum altitude reached by the rocket, we need to find the vertical displacement. We can do this by calculating the vertical velocity at the given time and then using the kinematic equation:
vf = vi + at
where:
vf = final vertical velocity
vi = initial vertical velocity
a = acceleration
t = time
Given:
vi = 103 m/s (initial speed)
a = -9.8 m/s^2 (acceleration due to gravity, negative because it acts in the opposite direction of the rocket's motion)
t = 3.00 s
Now, let's calculate the vertical velocity at this time:
vf = vi + at
vf = 103 m/s + (-9.8 m/s^2) * 3.00 s
vf = 103 m/s - 29.4 m/s
vf = 73.6 m/s
The vertical velocity at this time is 73.6 m/s.
Next, we can calculate the displacement using the equation:
Δy = vi * t + (1/2) * a * t^2
where:
Δy = displacement in the y direction (vertical displacement)
Plugging in the values, we get:
Δy = 103 m/s * 3.00 s + (1/2) * (-9.8 m/s^2) * (3.00 s)^2
Δy = 309 m + (1/2) * (-9.8 m/s^2) * 9.00 s^2
Δy = 309 m - 44.1 m
Δy = 264.9 m
The vertical displacement or maximum altitude reached by the rocket is 264.9 m.
(b) Find its total time of flight:
To find the total time of flight, we need to find the time it takes for the rocket to reach its maximum altitude and then come back down to the ground. Since the initial upward motion and the downward motion are symmetrical, the total time of flight is twice the time it takes to reach the maximum altitude.
The time it takes to reach the maximum altitude is given by:
t_max = vf / |a|
where:
t_max = time to reach maximum altitude
Plugging in the values, we get:
t_max = 73.6 m/s / 9.8 m/s^2
t_max = 7.51 s (rounded to two decimal places)
Since the upward and downward motions are symmetrical, the total time of flight is:
Total time of flight = 2 * t_max
Total time of flight = 2 * 7.51 s
Total time of flight = 15.02 s (rounded to two decimal places)
The total time of flight of the rocket is 15.02 s.
(c) Find its horizontal range:
To find the horizontal range, we need to calculate the horizontal displacement. We can do this using the horizontal velocity and the time of flight. The horizontal velocity remains constant throughout the motion.
The horizontal displacement or range is given by:
Δx = horizontal velocity * total time of flight
Given:
horizontal velocity = vi * cos(θ) (where θ is the launch angle, 56.0° in this case)
vi = 103 m/s (initial speed)
θ = 56.0° (launch angle)
total time of flight = 15.02 s (from part b)
Let's plug in the values and calculate the horizontal range:
Δx = 103 m/s * cos(56.0°) * 15.02 s
Δx = 103 m/s * 0.5591929 * 15.02 s
Δx = 859.5 m
The horizontal range of the rocket is 859.5 m.
So, to summarize:
(a) The maximum altitude reached by the rocket is 264.9 m.
(b) The total time of flight of the rocket is 15.02 s.
(c) The horizontal range of the rocket is 859.5 m.