A test rocket is launched by accelerating it along a 200.0-m incline at 1.25m/s2 starting from rest at point A (the figure (Figure 1) .) The incline rises at 35.0 ∘ 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 greatest horizontal range of the rocket beyond point A.

To find the greatest horizontal range of the rocket beyond point A, we need to first calculate the time it takes for the rocket to leave the inclined plane.

The acceleration along the incline is given as 1.25 m/s^2, and the angle of the incline is 35.0 degrees. We can use the component of gravity parallel to the incline to find the normal force and the frictional force acting on the rocket.

The component of gravity parallel to the incline is:
F_parallel = m * g * sin(theta)
F_parallel = m * 9.8 m/s^2 * sin(35.0 degrees)

Since the rocket starts from rest, the net force along the incline is equal to the force of acceleration:
F_net = F_parallel - f_friction
m * a = m * g * sin(theta) - f_friction

We know that the frictional force is given by:
f_friction = u * N
where u is the coefficient of friction and N is the normal force.

The normal force can be calculated using the component of gravity perpendicular to the incline:
F_perpendicular = m * g * cos(theta)
N = F_perpendicular
N = m * g * cos(35.0 degrees)

Substituting the value of N into the equation for frictional force, we get:
f_friction = u * m * g * cos(35.0 degrees)

Now, let's substitute this back into the net force equation and solve for the mass of the rocket:
m * a = m * g * sin(35.0 degrees) - u * m * g * cos(35.0 degrees)
a = g * sin(35.0 degrees) - u * g * cos(35.0 degrees)

We can solve for the mass of the rocket:
m = (g * sin(35.0 degrees)) / (a + u * g * cos(35.0 degrees))

Now, let's find the time it takes for the rocket to leave the incline. We can use the kinematic equation:
d = v0 * t + (1/2) * a * t^2

Since the rocket starts from rest, the initial velocity (v0) is 0. The distance (d) traveled along the incline is given as 200.0 m. The acceleration (a) is 1.25 m/s^2. We can rearrange the equation to solve for time (t):
t = sqrt((2 * d) / a)

Substituting the values, we get:
t = sqrt((2 * 200.0 m) / (1.25 m/s^2))

Now that we have the time it takes for the rocket to leave the incline, we can calculate the horizontal displacement. We know that once the rocket leaves the incline, it only experiences the force of gravity.

The horizontal displacement can be calculated using the equation:
x = v0 * t + (1/2) * a * t^2

Since the horizontal velocity (v0) is constant after leaving the incline, the equation simplifies to:
x = v0 * t

The initial horizontal velocity can be calculated using the vertical velocity right before leaving the incline, which is equal to the horizontal velocity along the incline.

The horizontal velocity along the incline can be calculated using the equation:
v0 = a * t

Substituting the values, we get:
v0 = 1.25 m/s^2 * sqrt((2 * 200.0 m) / (1.25 m/s^2))
v0 = 15.85 m/s

Finally, we can calculate the greatest horizontal range by multiplying the initial horizontal velocity by the time:
range = v0 * t
range = 15.85 m/s * sqrt((2 * 200.0 m) / (1.25 m/s^2))