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 determine the rocket's velocity when it leaves the incline and calculate the horizontal distance it covers after leaving the incline.
Step 1: Find the velocity of the rocket when it leaves the incline:
We can use the kinematic equation for velocity to find the velocity of the rocket when it leaves the incline. The equation is:
v^2 = u^2 + 2as
where:
v = final velocity,
u = initial velocity,
a = acceleration,
s = displacement.
Since the rocket starts from rest, its initial velocity (u) is 0 m/s. The displacement (s) is the distance along the incline, which is given as 200.0 m. The acceleration (a) is the acceleration along the incline, which is given as 1.25 m/s^2. Plugging in these values into the kinematic equation, we can solve for v:
v^2 = 0^2 + 2(1.25 m/s^2)(200.0 m)
v^2 = 2.50 m^2/s^2 * 200.0 m
v^2 = 500 m^2/s^2
v ≈ 22.36 m/s
Therefore, the velocity of the rocket when it leaves the incline is approximately 22.36 m/s.
Step 2: Calculate the horizontal distance covered by the rocket after leaving the incline:
Since the rocket is subject only to gravity after leaving the incline, its horizontal motion will be governed by the principle of uniform motion. The horizontal distance covered (range) can be calculated using the equation:
range = horizontal velocity * time
To calculate the horizontal velocity, we need to resolve the velocity of the rocket into its horizontal component. The horizontal component of velocity is given by:
horizontal velocity = velocity * cos(theta)
where theta (θ) is the angle of the incline with respect to the horizontal, which is given as 35.0 degrees. Plugging in the values, we can calculate the horizontal velocity:
horizontal velocity = 22.36 m/s * cos(35.0 degrees)
horizontal velocity ≈ 18.19 m/s
Now, we need to calculate the time it takes for the rocket to reach the ground. Since the vertical motion is influenced only by gravity, we can use the equation:
s = ut + (1/2)at^2
where:
s = displacement in the vertical direction,
u = initial vertical velocity,
a = vertical acceleration (due to gravity),
t = time.
The vertical displacement (s) is the distance the rocket has to fall from the point it leaves the incline to the ground, which is the same as the height of the incline at that point. The vertical acceleration (a) due to gravity is approximately 9.8 m/s^2. The initial vertical velocity (u) is the vertical component of the velocity of the rocket just before it leaves the incline, and it can be calculated using:
initial vertical velocity = velocity * sin(theta)
Plugging in the values, we can calculate the initial vertical velocity:
initial vertical velocity = 22.36 m/s * sin(35.0 degrees)
initial vertical velocity ≈ 12.86 m/s
The vertical displacement is equal to the height of the incline at point A, which is given in the figure.
Now, we can calculate the time (t) it takes for the rocket to reach the ground:
s = ut + (1/2)at^2
200.0 m = 12.86 m/s * t + (1/2)(-9.8 m/s^2)(t^2)
200.0 m = 12.86 m/s * t - 4.9 m/s^2 * t^2
This is a quadratic equation. We can solve for t by rearranging the equation and applying the quadratic formula:
4.9t^2 - 12.86t + 200.0 = 0
Using the quadratic formula, the solutions for t are approximately t = 4.78 s and t = 8.41 s. Since we're interested in the time it takes for the rocket to reach the ground, we use the positive value, which is t = 8.41 s.
Finally, we can calculate the horizontal distance covered by the rocket after leaving the incline:
range = horizontal velocity * time
range = 18.19 m/s * 8.41 s
range ≈ 152.85 m
Therefore, the greatest horizontal range of the rocket beyond point A is approximately 152.85 meters.
To find the greatest horizontal range of the rocket beyond point A, we need to determine the time it takes for the rocket to leave the incline and the horizontal component of its velocity at that point.
First, let's find the time it takes for the rocket to leave the incline. We can do this by using the kinematic equation:
\[v_f = v_i + at\]
where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
From the given information, the initial velocity (\(v_i\)) is 0 m/s, the acceleration (\(a\)) is 1.25 m/s\(^2\), and the final velocity (\(v_f\)) at the moment the rocket leaves the incline is unknown.
Since the rocket starts from rest at point A, we can assume that the velocity at that point is entirely due to the acceleration along the incline. Therefore, the final velocity (\(v_f\)) can be calculated using the equation:
\[v_f = v_i + at\]
\[v_f = 0 + (1.25 \times t)\]
where \(t\) is the time.
We can also find the vertical component of the velocity at the moment the rocket leaves the incline using the equation:
\[v_f^2 = v_i^2 + 2gh\]
where \(g\) is the acceleration due to gravity and \(h\) is the height of the incline.
Since the rocket starts from rest at point A, the initial velocity (\(v_i\)) is 0 m/s, and the height (\(h\)) of the incline is given as 200 m.
Substituting these values into the equation, we get:
\[(1.25t)^2 = 0 + 2 \times 9.8 \times 200\]
Solving this equation will give us the time it takes for the rocket to leave the incline (\(t\)).
Once we find the time (\(t\)), we can calculate the horizontal component of the velocity at the moment the rocket leaves the incline using the equation:
\[v_{xf} = v_{xi} + a_x \times t\]
where \(v_{xf}\) is the final horizontal velocity, \(v_{xi}\) is the initial horizontal velocity, \(a_x\) is the horizontal acceleration, and \(t\) is the time.
Since the rocket starts from rest at point A, the initial horizontal velocity (\(v_{xi}\)) is 0 m/s, the horizontal acceleration (\(a_x\)) is also 0 m/s\(^2\) because the rocket is subject only to gravity after leaving the incline.
Therefore, the final horizontal velocity (\(v_{xf}\)) is just the horizontal component of the velocity at the moment the rocket leaves the incline, which can be calculated as:
\[v_{xf} = 1.25 \times t \times \cos(35.0^\circ)\]
Finally, to find the greatest horizontal range, we can use the equation:
\[R = \frac{v_{xf}^2}{g}\]
where \(R\) is the range and \(g\) is the acceleration due to gravity.
By substituting the calculated values for \(v_{xf}\) and \(g\) into the equation, we can determine the greatest horizontal range of the rocket beyond point A.