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
I've figured out the upward motion part of the problem still having trouble with the downward motion (landing angle and final velocity). Can someone explain thanks a lot for all the help
I did the entire thing for you.
Vf is the vector addition of Vhoriz, Vvertical.
Vhoriz= 15cosTheta.
Vvertical= 15sinTheta - 4.9*3.2
Now the landing angle: Tan Thetaland=Vv/Vh and I just laid out those two for you.
Vlanding= sqrt(Vv^2 + Vh^2) or you can do it with trig.
To determine the downward motion of the daredevil's motorcycle during landing, we need to calculate the landing angle and final velocity.
1. First, let's calculate the initial vertical velocity (upward motion) using the given information. We know that the motorcycle's flight time is 3.2 seconds and the vertical distance (drop) is 15 meters.
The formula to calculate vertical displacement is:
Vertical displacement (drop) = Initial vertical velocity * Time + 0.5 * Acceleration * Time^2
Since the motorcycle starts from rest vertically, the initial vertical velocity is 0. Therefore, the equation simplifies to:
15 m = 0 * 3.2 s + 0.5 * Acceleration * (3.2 s)^2
We can rewrite the equation to solve for acceleration (a):
15 m * 2 / (3.2 s)^2 = Acceleration
Simplifying, we get:
Acceleration = 2.34 m/s^2
2. Now, let's calculate the horizontal distance the daredevil needs to travel. The planned distance was 60 m, but we know the actual distance is 69.6 m.
The horizontal distance (displacement) can be calculated using the formula:
Horizontal displacement = Initial horizontal velocity * Time
Since the daredevil wants the same flight time for both upward and downward motions, and there is no horizontal force acting on the motorcycle, the initial horizontal velocity stays constant throughout the entire jump.
Therefore, we can calculate it by dividing the horizontal distance by the flight time:
Initial horizontal velocity = 69.6 m / 3.2 s = 21.75 m/s
3. Next, we need to calculate the landing angle. The landing angle (θ) can be determined using trigonometry.
The vertical velocity (downward motion) can be found using the formula:
Vertical velocity = Initial vertical velocity + Acceleration * Time
Since the motorcycle is in freefall during the downward motion, there is no initial vertical velocity (it is 0). Therefore, the vertical velocity (Vv) is simply the acceleration multiplied by the flight time:
Vertical velocity = 2.34 m/s^2 * 3.2 s = 7.488 m/s
The horizontal velocity (Vh) remains constant throughout, as there is no horizontal force acting on the motorcycle. The initial horizontal velocity we calculated earlier is the horizontal velocity during the entire jump:
Horizontal velocity = 21.75 m/s
Now, using trigonometry, we can calculate the landing angle:
θ = arctan(Vertical velocity / Horizontal velocity)
θ = arctan(7.488 m/s / 21.75 m/s) ≈ 19.76 degrees
4. Finally, let's calculate the final velocity upon landing. The final velocity (Vf) can be found using the equation:
Vf = sqrt((Vertical velocity)^2 + (Horizontal velocity)^2)
Vf = sqrt((7.488 m/s)^2 + (21.75 m/s)^2) ≈ 23.08 m/s
To summarize:
- Landing angle (θ) ≈ 19.76 degrees
- Final velocity (Vf) ≈ 23.08 m/s
Please note that these calculations assume ideal conditions and neglect factors such as air resistance.