A Hollywood daredevil plans to jump the canyon shown in the figure 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 65.7 m across. If he desires a 3.1-second flight time, what is the correct angle for his launch ramp (deg)? What is the correct angle for his landing ramp (give a positive angle below the horizontal)? is his predicted landing velocity. (Neglect air resistance.)
To find the correct angle for the launch ramp, we can use the projectile motion equations. Let's start by finding the initial velocity (v0) of the motorcycle.
Given:
Drop height (h) = 15 m
Flight time (t) = 3.1 s
Since there is no horizontal acceleration (neglecting air resistance), the time taken to reach the highest point (the peak of the trajectory) will be half of the total flight time.
t_peak = t / 2 = 3.1 s / 2 = 1.55 s
To calculate the initial vertical velocity (v0y) at the peak, we can use the following equation:
v0y = g * t_peak
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
v0y = 9.8 m/s^2 * 1.55 s = 15.19 m/s
Now, let's calculate the horizontal distance (x) the motorcycle needs to cover.
Given:
Original planned distance (x_p) = 60 m
Actual distance (x_a) = 65.7 m
To find the correct angle for the launch ramp, we'll compare the horizontal distances covered with the original and actual distances.
x_p = v0 * t
Solving for v0:
v0 = x_p / t
v0 = 60 m / 3.1 s = 19.35 m/s
Now that we have the initial vertical velocity (v0y) and the initial horizontal velocity (v0), we can calculate the launch ramp angle (θ) using the following trigonometric equation:
tan(θ) = v0y / v0
θ = arctan(v0y / v0)
θ = arctan(15.19 m/s / 19.35 m/s) = 37.21 degrees (approximately)
Therefore, the correct angle for the launch ramp is approximately 37.21 degrees.
To find the correct angle for the landing ramp, we need to calculate the predicted landing velocity (v_landing) using the conservation of energy principle. The potential energy at the peak of the trajectory will be converted to kinetic energy at the bottom of the trajectory (neglecting air resistance).
Potential energy at the peak = m * g * h
Kinetic energy at the bottom = 1/2 * m * v_landing^2
Since the mass (m) of the motorcycle cancels out, we can equate the potential energy to the kinetic energy:
m * g * h = 1/2 * m * v_landing^2
Simplifying:
g * h = 1/2 * v_landing^2
v_landing^2 = 2 * g * h
v_landing = √(2 * 9.8 m/s^2 * 15 m) = √(294 m^2/s^2) = √294 m/s ≈ 17.15 m/s
Now, let's calculate the angle for the landing ramp (θ_landing) using the trigonometric equation:
tan(θ_landing) = v_landing / v0
θ_landing = arctan(v_landing / v0)
θ_landing = arctan(17.15 m/s / 19.35 m/s) = 41.73 degrees (approximately)
Therefore, the correct angle for the landing ramp (below the horizontal) is approximately 41.73 degrees.
Please note that these calculations neglect air resistance and assume ideal conditions. In reality, air resistance and other factors may affect the actual flight and landing of the motorcycle.