Helped needed with projectile motion given only angle and x/y distance

In summary, the conversation discusses a physics professor who performs daredevil stunts in his spare time. His latest stunt involved attempting to jump across a river on a motorcycle, with specific measurements for the takeoff ramp, width of the river, and elevation of the far bank. The conversation also includes two questions about the necessary speed at the top of the ramp and where he would land if his speed was only half the required amount. The speaker mentions struggling to solve the problem and asks for help and clarification on the method of solving it.
  • #1
TheLunchGun
2
0
1. A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle. The takeoff ramp was inclined at 53.0 degrees, the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of the ramp. The river itself was 100 m below the ramp. You can ignore air resistance.

a. What should his speed have been at the top of the ramp for him to have just made it to the edge of the far bank?

b. If his speed was only half the value found in (a), where did he land?


This problem has been a thorn in my side for many an hour. Some help would be great if anyone has some time.
 
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  • #2
What work have you done so far? Where are you getting stuck?
 
  • #3
I've tried a number of different angles on it but nothing seems to go anywhere. I actually already have the answers, but I want to know the method of solving it. I can post them if needed.
 

Related to Helped needed with projectile motion given only angle and x/y distance

1. How do you calculate the initial velocity of a projectile with only the angle and x/y distance?

The initial velocity of a projectile can be calculated using the following formula: V0 = (x / cosθ) * √(g/2y), where V0 is the initial velocity, x is the horizontal distance, θ is the angle of projection, g is the acceleration due to gravity, and y is the vertical distance.

2. Why is the angle of projection important in projectile motion?

The angle of projection plays a crucial role in determining the trajectory of a projectile. It affects the horizontal and vertical components of the initial velocity, which in turn determine the shape and range of the projectile's path.

3. Can you solve for the final velocity of a projectile with only the angle and x/y distance?

Yes, the final velocity of a projectile can be calculated using the following formula: V = √(V02 + 2gy), where V is the final velocity, V0 is the initial velocity, g is the acceleration due to gravity, and y is the vertical distance.

4. How do you determine the maximum height of a projectile with only the angle and x/y distance?

The maximum height of a projectile can be calculated using the following formula: H = (V02 * sin2θ) / 2g, where H is the maximum height, V0 is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

5. Is it possible to predict the landing point of a projectile with only the angle and x/y distance?

Yes, the landing point of a projectile can be predicted by using the equations of motion to calculate the time of flight and the horizontal distance traveled. However, factors such as air resistance and wind may affect the actual landing point.

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