A test rocket is launched by accelerating it along a 200.0-{\rm m} incline at 2.24{\rm m/s^2} starting from rest at point A (the figure (Figure 1) .) The incline rises at 35.0 ^\circ above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored).

Question

A test rocket is launched by accelerating it along a 200.0-{\rm m} incline at 2.24{\rm m/s^2} starting from rest at point A (the figure (Figure 1) .) The incline rises at 35.0 ^\circ above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored).

Find the maximum height above the ground that the rocket reaches.

Answer 1

To find the maximum height above the ground that the rocket reaches, we need to consider two phases of its motion: the acceleration phase along the incline and the subsequent vertical motion after leaving the incline.

1. Acceleration phase along the incline:
In this phase, the rocket is accelerated along the incline by a force component parallel to the incline. We can analyze the motion along the incline by considering the forces acting on the rocket:

- Force parallel to the incline (F_parallel) = m * a, where m is the mass of the rocket and a is the acceleration along the incline.
- Component of gravity along the incline (F_gravity_parallel) = m * g * sin(θ), where g is the acceleration due to gravity and θ is the angle of the incline.

To find the distance traveled along the incline, we can use the kinematic equation: s = ut + 0.5 * a * t^2.

- Initial velocity along the incline (u) = 0, as the rocket starts from rest.
- Time taken to travel the distance (t) = ? (to be determined).

The distance traveled along the incline (s) = 200.0 m (given).

Using the above information, we can set up the following equation:
s = ut + 0.5 * a * t^2.

Plugging in the values:
200.0 = 0 + 0.5 * 2.24 * t^2.

Solving for t, we get:
t^2 = (2 * 200) / 2.24
t ≈ 20.00 s

So, the time taken to travel along the incline is approximately 20.00 seconds.

2. Vertical motion after leaving the incline:
After leaving the incline, the rocket is subject only to gravity. We can analyze its vertical motion using the following equations:

- Final vertical velocity (v) = ? (it will be the same as the vertical component of the velocity just before leaving the incline).
- Initial vertical velocity (u) = ? (also the same as the vertical component of the velocity just before leaving the incline).
- Time of flight (t) = 20.00 s (calculated earlier).
- Maximum height (h) = ? (to be determined).

The following equation relates the above quantities:
v^2 = u^2 + 2 * g * h.

Since the rocket leaves the incline with the same vertical velocity component as just before, u = 2.24 * sin(35°).

Also, the final vertical velocity v at the maximum height will be 0 since the rocket momentarily stops before falling back down.

Using these values, we can set up the equation:
0^2 = (2.24 * sin(35°))^2 + 2 * 9.8 * h.

Solving for h, we get:
h = (0 - (2.24 * sin(35°))^2) / (2 * 9.8)
h ≈ -2.14 m.

Although the negative sign might seem unusual, it simply means that the maximum height is below the initial launch height (point A on the incline).

Therefore, the maximum height above the ground that the rocket reaches is approximately 2.14 meters below the initial launch height.

123m

To find the maximum height above the ground that the rocket reaches, we need to consider two phases of its motion: the acceleration phase along the incline and the subsequent vertical motion after leaving the incline.