Work/Energy Problem: Maximum Height of Daredevil on Motorcycle

[myelevatorbeat]

Homework Statement

A daredevil on a motorcycle leaves the end of a ramp with a speed of 39.0 m/s as in Figure P5.23. If his speed is 36.8 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.

The Attempt at a Solution

I figured I would start by finding the height when he is going off the ramp so I did this:

V^2=Vo^2-2g(Y-Yo)
39^2=0^2-2(9.80)(Y-Yo)
and wound up with
(Y-Yo)=-77.6 m

I'm not sure if I'm doing this correctly because I got a negative height. Should I just get rid of the negative sign and use this as my height?

 

[Dick]

You shouldn't just get rid of the negative sign, because the number is not right either. You didn't even use 36.8m/sec in your solution. What makes you think it doesn't matter?

 

[Dick]

Correct solution is as follows : The initial velocity Vo (36.8 m/s) can be resolved into x and y components : Vo(x) = Vo cos (theta) and Vo(y) = Vo sin (theta) where theta is the angle made by the ramp with the horizontal. At the highest point, only the verticle component of velocity becomes zero (Vo(x) remains constant throughout). Substituting V = 0, Vo = Vo sin(theta), a = -g in V^2 - Vo^2 = 2a(Y - Yo) we get

-Vo^2 [sin(theta)]^2 = -2g (Y - Yo)

(Y - Yo) = Vo^2 [sin(theta)]^2/2g

Hope this helps.

You don't need to resolve it into components. In fact, I think this is where the OP went wrong. In this problem the equation being used expresses conservation of energy. The energies only depend on the difference in the magnitudes of the velocities and the vertical displacement.

 

[Vijay Bhatnagar]

The initial velocity Vo (39 m/s) can be resolved into x and y components : Vox = Vo cos (theta) and Voy = Vo sin (theta) where theta is the angle made by the ramp with the horizontal. At the highest point, the verticle component of velocity becomes zero. The horizontal component Vox remains constant throughout as there is no force acting on the motor cyclist in that direction. Thus, at the highest point Vox = Vo cos(theta) = 36.8 m/s. Solving get theta. Substituting Vy = 0, Vo(y) = Vo sin(theta), a = -g in Vy^2 - Voy^2 = 2a(Y - Yo) solve for (Y - Yo).

Hope this helps.

 

[Dick]

The initial velocity Vo (39 m/s) can be resolved into x and y components : Vox = Vo cos (theta) and Voy = Vo sin (theta) where theta is the angle made by the ramp with the horizontal. At the highest point, the verticle component of velocity becomes zero. The horizontal component Vox remains constant throughout as there is no force acting on the motor cyclist in that direction. Thus, at the highest point Vox = Vo cos(theta) = 36.8 m/s. Solving get theta. Substituting Vy = 0, Vo(y) = Vo sin(theta), a = -g in Vy^2 - Voy^2 = 2a(Y - Yo) solve for (Y - Yo).

Hope this helps.

You are correct that you CAN do it that way. But you can also do it without solving for theta using the formula presented by the OP. The only problem is that the OP is not using it right.

 

[Vijay Bhatnagar]

You don't need to resolve it into components. In fact, I think this is where the OP went wrong. In this problem the equation being used expresses conservation of energy. The energies only depend on the difference in the magnitudes of the velocities and the vertical displacement.

This can be an alternative approach : KE at the highest point - KE at the end of the ramp = Change in PE . I agree this is the better approach.