A Hollywood Daredevil plans to jump a canyon on a motorcycle. There is a 15 m drop and the horizontal distance originally planned was 60 m but it turns out the canyon is really 69.6 m across. If he desires a a 3.2 second flight time
a. what is the correct angle of his launch ramp?
b.what is his correct launch speed?
c.what is the correct angle for the landing ramp?
d.what is his predicted landing velocity?
a. The 3.2 s flight time tells you what the vertical launch component, Vy, must be.
Vy*3.2 s - g*(3.2)^2/2 = -15
Vy = 7.87 m/s
The width of the canyon and the flight time tell you what the horizontal velocity component, Vx, must be
Vx*3.2s = 69.6 m
Vx = 21.75 m/s
The ratio Vy/Vx is the tangent of the launch angle.
b. V = sqrt (Vx^2 + Vy^2)
c. Calculate the final vertical velocity at landing. It will be larger than the launch value of Vy because of the loss of height. Vy(final)/Vx will be the landing ramp angle tangent.
d. You can get this easily from the launch velocity and the 15 meter drop in altitude. V^2 increases by 2 g H, where H = 15 m.
To calculate the correct angle of the launch ramp, launch speed, landing ramp angle, and predicted landing velocity for the Hollywood Daredevil's motorcycle canyon jump, we can use the principles of projectile motion.
a. Angle of Launch Ramp:
To find the correct angle of the launch ramp, we'll first calculate the time it takes for the motorcycle to reach the peak of its trajectory using the flight time.
1. Find the time to reach the peak:
Using the equation of motion, h = v₀²sin²(θ) / (2g), where h is the vertical distance, v₀ is the initial vertical velocity, θ is the angle of the launch ramp, and g is the acceleration due to gravity:
15 m = v₀²sin²(θ) / (2 * 9.8 m/s²)
v₀²sin²(θ) = 294 m²/s²
(sin θ)² = 294 m²/s² / (v₀² * 2 * 9.8 m/s²)
(sin θ)² = 15 m² / (v₀² * 9.8 m/s²)
2. Find the time of flight:
Since the total flight time is given as 3.2 seconds, the time to reach the peak will be half of the total flight time:
t_peak = 3.2 s / 2 = 1.6 s
3. Substitute the values:
Using the value of t_peak, we can solve for (sin θ)²:
(sin θ)² = 15 m² / (v₀² * 9.8 m/s²)
b. Launch Speed:
To find the correct launch speed, we can use the horizontal distance and the time of flight.
1. Calculate the horizontal velocity:
The horizontal velocity is given by the equation, v_x = d / t, where v_x is the horizontal velocity, d is the horizontal distance, and t is the time of flight:
v_x = 69.6 m / 3.2 s
2. Calculate the launch speed:
The launch speed can be calculated using the horizontal velocity and the launch ramp angle, θ:
v₀ = v_x / cos(θ)
c. Angle of Landing Ramp:
To calculate the correct angle of the landing ramp, we need to consider the horizontal distance and the total flight time.
1. Calculate the horizontal velocity:
Using the same formula as above, v_x = d / t:
v_x = 69.6 m / 3.2 s
2. Calculate the angle of the landing ramp:
The angle of the landing ramp, θ', can be calculated using the horizontal velocity and the launch angle (θ):
tan(θ') = v_x / (v₀ * cos(θ))
d. Predicted Landing Velocity:
To find the predicted landing velocity, we can calculate the vertical velocity at the time of landing.
1. Calculate the vertical velocity:
Using the equation of motion, v_y = v₀ * sin(θ) - g * t:
v_y = v₀ * sin(θ) - 9.8 m/s² * 3.2 s
2. Calculate the predicted landing velocity:
The predicted landing velocity, V_landing, is the magnitude of the vertical velocity at landing:
V_landing = sqrt((v_y)² + (v_x)²)
Note: The predicted values above are based on a simplified model of projectile motion and do not account for factors such as air resistance and other real-world complexities.
Please provide the launch ramp angle (θ) and we can calculate the other values accordingly.
To answer these questions, we can use the principles of projectile motion. Projectile motion refers to the motion of an object in the air under the influence of gravity. In this case, the daredevil's motorcycle can be treated as a projectile.
a. To determine the correct angle of the launch ramp, we can use the horizontal and vertical components of motion. Let's start by finding the time it takes for the daredevil to reach the maximum height of the jump.
Given:
Vertical displacement, Δy = -15 m (negative sign because the motorcycle is dropping)
Flight time, t = 3.2 s
Initial vertical velocity, Vy_initial = 0 m/s (as the motorcycle starts from the ground)
Using the equation:
Δy = Vy_initial * t + (1/2) * g * t^2
Substituting the given values, we can solve for the time:
-15 = 0 * 3.2 + (1/2) * (-9.8) * (3.2)^2
Solving this equation, we get t ≈ 1.59 s.
Next, we need to find the horizontal displacement, Δx, during this time:
Given:
Horizontal displacement, Δx = 69.6 - 60 = 9.6 m (as the canyon is wider than initially planned)
Using the equation:
Δx = Vx * t
Solving for Vx:
9.6 = Vx * 1.59
Vx = 9.6 / 1.59 ≈ 6.04 m/s
Now, to find the angle of the launch ramp, we can use the equation for the horizontal and vertical velocities:
Vx = V_initial * cos(θ)
Vy_initial = V_initial * sin(θ)
Substituting the known values, we have:
6.04 = V_initial * cos(θ)
0 = V_initial * sin(θ)
From the second equation, we can conclude that sin(θ) = 0, which means the angle θ must be 0° or 180° in radians. However, for a daredevil jump, θ should be in the range of 0° to 90°. Therefore, the correct angle for the launch ramp is 0°.
b. To find the correct launch speed, we can use the horizontal component of the velocity:
Vx = V_initial * cos(θ)
Substituting the known values, we have:
6.04 = V_initial * cos(0)
Since cos(0) = 1, we can solve for V_initial:
V_initial = 6.04 m/s
Therefore, the correct launch speed is 6.04 m/s.
c. To find the correct angle for the landing ramp, we need to consider the vertical component of the motion. At the end of the jump, the daredevil should land at the same height he started from (ground level).
Using the equation:
Δy = Vy_initial * t + (1/2) * g * t^2
Substituting the given values:
0 = 0 * 3.2 + (1/2) * (-9.8) * (3.2)^2
Solving this equation, we find t ≈ 1.59 s.
Now, we can use the equation for the vertical component of the velocity:
Vy_final = Vy_initial + g * t
Given that Vy_initial = 0 and g = -9.8 m/s², we have:
Vy_final = -9.8 * 1.59 ≈ -15.58 m/s
As the daredevil lands on the same height he started from, the vertical component of the velocity must be the same but in the opposite direction. Therefore, the correct angle for the landing ramp is 180° in radians.
d. To find the predicted landing velocity, we can use the equation for the total velocity:
V_final = sqrt(Vx^2 + Vy^2)
Substituting the known values:
V_final = sqrt((6.04 m/s)^2 + (-15.58 m/s)^2)
≈ sqrt(36.49 + 242.74)
≈ sqrt(279.23)
≈ 16.71 m/s
Therefore, the predicted landing velocity is approximately 16.71 m/s.