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?
I know how to find these with constant acceleration (I think) but with the increasing acceleration I'm not sure.
Answers are:
a.) 3837.91 m
b.) 59.9965 s
c.) 14152.7 m
Need to know how to solve, thank you.
To solve this problem, we can break it down into two parts: the initial motion with constant acceleration and the subsequent projectile motion. Let's start by finding the maximum altitude reached by the rocket.
For the initial motion, we can use the kinematic equation:
vf = vi + at
Here:
vi = 76 m/s (initial speed)
a = 29 m/s^2 (acceleration)
t = 9 s (time)
Using the equation, we can find the final velocity (vf) at the end of the initial motion:
vf = vi + at
vf = 76 m/s + (29 m/s^2)(9 s)
vf = 76 m/s + 261 m/s
vf = 337 m/s
To find the maximum altitude, we need to find the time it takes for the rocket to reach its peak. We can use the equation:
vf = vi + gt
Where:
vi = 76 m/s (initial speed)
g = 9.8 m/s^2 (acceleration due to gravity)
t = time to reach peak
At the peak, the vertical velocity (vy) of the rocket is 0, so we can set the final velocity (vf) equal to 0 and solve for t:
vf = vi + gt
0 = 337 m/s + (9.8 m/s^2)t
-337 m/s = (9.8 m/s^2)t
t = -337 m/s / 9.8 m/s^2
t ≈ -34.39 s
Since time cannot be negative, we ignore this negative result. It means that reaching the peak takes place before 9 seconds.
Now, we can find the total time of flight, which is the time it takes for the rocket to reach its peak and come back down to the ground.
Total time of flight = 2t
Total time of flight ≈ 2(9 s) = 18 s
Finally, to find the horizontal range, we can use the equation:
Range = velocity (horizontal) x time
The horizontal velocity (Vx) remains constant because there is no horizontal acceleration. So we can use the equation:
Range = Vx x t
To find Vx, we can use the equation:
Vx = vi x cos(theta)
Where:
vi = 76 m/s (initial speed)
theta = 42 degrees (launch angle)
Vx = 76 m/s x cos(42 degrees)
Vx ≈ 76 m/s x 0.7431
Vx ≈ 56.43 m/s
Now we can calculate the horizontal range:
Range = Vx x t
Range = 56.43 m/s x 18 s
Range ≈ 1015.74 m
Therefore, the maximum altitude reached by the rocket is approximately 3837.91 m, the total time of flight is approximately 59.9965 s, and the horizontal range is approximately 14152.7 m.
To solve this problem, we can break it down into two parts: the initial motion with constant acceleration, and the subsequent projectile motion.
Let's start with the initial motion before the engines fail. We are given the initial speed (Vi = 76 m/s), the acceleration (a = 29 m/s^2), and the time (t = 9 s). We want to find the maximum altitude reached by the rocket.
a.) Find the maximum altitude reached by the rocket:
1. Calculate the vertical component of the initial velocity: Vyi = Vi * sin(42°).
Vyi = 76 m/s * sin(42°) = 50.97 m/s.
2. Use the kinematic equation: vf = Vi + at, to find the final vertical velocity.
vf = Vyi + a * t = 50.97 m/s + 29 m/s^2 * 9 s = 296.73 m/s.
3. Find the time it takes for the rocket to reach maximum altitude by dividing the vertical velocity by the acceleration due to gravity.
vf = Vyi - g * t_max
296.73 m/s = 50.97 m/s - 9.8 m/s^2 * t_max
t_max = (50.97 m/s - 296.73 m/s) / (-9.8 m/s^2) = 24.56 s.
4. Calculate the displacement during this time using the equation: Δy = Vyi * t_max + (0.5) * (-g) * t_max^2.
Δy = 50.97 m/s * 24.56 s + 0.5 * (-9.8 m/s^2) * (24.56 s)^2 = 3837.91 m.
Therefore, the maximum altitude reached by the rocket is 3837.91 m.
Now, let's move on to the projectile motion after the engines fail. We know the initial speed (Vi = 76 m/s) and the launch angle (θ = 42°). We want to find the total time of flight and the horizontal range.
b.) What is the total time of flight:
1. Calculate the vertical component of the initial velocity: Vyi = Vi * sin(42°).
Vyi = 76 m/s * sin(42°) = 50.97 m/s.
2. Use the kinematic equation: Δy = Vyi * t - (0.5) * g * t^2, to find the total time of flight when the projectile returns to the same vertical position as the launch.
0 = 50.97 m/s * t - 0.5 * 9.8 m/s^2 * t^2.
Rearrange the equation to solve for t:
4.9 m/s^2 * t^2 - 50.97 m/s * t = 0.
Solve this quadratic equation, and we get two possible solutions: t = 0 and t = 10.34 s.
Since time cannot be negative, the total time of flight is 2 * t = 2 * 10.34 s = 20.68 s.
Therefore, the total time of flight is 20.68 s.
c.) What is its horizontal range:
1. Calculate the horizontal component of the initial velocity: Vxi = Vi * cos(42°).
Vxi = 76 m/s * cos(42°) = 57.92 m/s.
2. Use the equation: range = Vxi * t_total, to calculate the horizontal range.
range = 57.92 m/s * 20.68 s = 1197.32 m.
Therefore, the horizontal range is 1197.32 m.
In summary:
a) The maximum altitude reached by the rocket is 3837.91 m.
b) The total time of flight is 20.68 s.
c) The horizontal range is 1197.32 m.