A rocket is launched at an angle of 56.0° above the horizontal with an initial speed of 96 m/s. The rocket 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 projectile.
(a) Find the maximum altitude reached by the rocket.(in meters)
(b) Find its total time of flight.(in seconds)
(c) Find its horizontal range. (in meters)
To find the maximum altitude reached by the rocket, we can use the kinematic equation for vertical displacement:
Δy = v₀y * t - (1/2) * g * t²
Here, Δy represents the vertical displacement (maximum altitude), v₀y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s²).
(a) The initial vertical velocity, v₀y, can be found using trigonometry. We know the initial speed of the rocket and the launch angle.
v₀y = v₀ * sin(θ)
Substituting the given values:
v₀y = 96 m/s * sin(56.0°)
Now, we can plug this value into the equation for vertical displacement:
Δy = (96 m/s * sin(56.0°)) * 3.00 s - (1/2) * (9.8 m/s²) * (3.00 s)²
Simplifying the equation will give us the maximum altitude reached by the rocket.
(b) To find the total time of flight, we can use the kinematic equation for time:
t = (v₀y - v₀y_point_of_impact) / g
Here, v₀y_point_of_impact is the vertical velocity at the point of impact (when the rocket lands), which is equal to -v₀y because the rocket lands at the same height from which it was launched.
Substituting the known values:
t = (v₀y - (-v₀y)) / g
Simplifying the equation will give us the total time of flight.
(c) To find the horizontal range, we can use the kinematic equation for horizontal displacement:
Δx = v₀x * t
Here, Δx represents the horizontal range, v₀x is the initial horizontal velocity, and t is the time.
The initial horizontal velocity, v₀x, can be found using trigonometry:
v₀x = v₀ * cos(θ)
Substituting the known values:
v₀x = 96 m/s * cos(56.0°)
Now, we can plug this value into the equation for horizontal displacement to find the range.
Remember to carry out the necessary calculations and unit conversions to get the final answers.
To solve this problem, we need to break it down into different parts.
(a) To find the maximum altitude reached by the rocket, we can use the kinematic equation for vertical motion. The rocket moves upward for 3.00 seconds, so we can use the following equation:
vf = vi + at
where vf is the final vertical velocity, vi is the initial vertical velocity, a is the vertical acceleration, and t is the time.
Initially, the vertical velocity of the rocket is vi = 96 m/s (sin 56°), because the rocket is launched at an angle of 56.0° above the horizontal. The vertical acceleration, a, is -9.8 m/s^2 (gravity acts downward). The time, t, is 3.00 s.
Using these values, we can calculate the final vertical velocity:
vf = vi + at
vf = (96 m/s (sin 56°)) + (-9.8 m/s^2)(3.00 s)
vf ≈ 96 m/s (0.829) - 29.4 m/s
vf ≈ 79.744 m/s - 29.4 m/s
vf ≈ 50.344 m/s
Now, we can use the kinematic equation for vertical motion to find the maximum altitude (hmax) reached by the rocket:
vf^2 = vi^2 + 2ahmax
Rearranging the equation, we get:
hmax = (vf^2 - vi^2) / (2a)
hmax = (50.344 m/s)^2 - (96 m/s (sin 56°))^2 / (2(-9.8 m/s^2))
hmax = 2534.97 m^2/s^2 - 7004.71 m^2/s^2 / -19.6 m/s^2
hmax ≈ -4469.74 m^2 / -19.6 m/s^2
hmax ≈ 228.12 m
Therefore, the maximum altitude reached by the rocket is approximately 228.12 meters.
(b) To find the total time of flight, we need to calculate the time it takes for the rocket to reach the maximum altitude, and then double that time.
The time it takes for the rocket to reach the maximum altitude is given by:
tmax = (vf - vi) / a
where tmax is the time to reach the maximum altitude, vi is the initial vertical velocity, vf is the final vertical velocity, and a is the vertical acceleration (which is -9.8 m/s^2).
Using the values we have already calculated:
tmax = (50.344 m/s - 96 m/s (sin 56°)) / -9.8 m/s^2
tmax = -45.756 m/s / -9.8 m/s^2
tmax ≈ 4.669 s
The total time of flight is then twice this value:
Total Time of Flight = 2 * tmax
Total Time of Flight ≈ 2 * 4.669 s
Total Time of Flight ≈ 9.338 s
Therefore, the total time of flight for the rocket is approximately 9.338 seconds.
(c) To find the horizontal range, we can use the horizontal velocity of the rocket.
The horizontal velocity can be found using the initial velocity and the launch angle:
Horizontal velocity (Vx) = vi * cosθ
where vi is the initial velocity (96 m/s) and θ is the launch angle (56.0°).
Vx = 96 m/s * cos 56.0°
Vx ≈ 96 m/s * 0.559
Vx ≈ 53.664 m/s
Now, we can find the horizontal range (R) using the equation:
Range (R) = Vx * t
where Vx is the horizontal velocity and t is the total time of flight.
Range (R) ≈ 53.664 m/s * 9.338 s
Range (R) ≈ 500.956 meters
Therefore, the horizontal range of the rocket is approximately 500.956 meters.