# Projectile Motion - Ramp Angle

1. Jan 12, 2008

### patv

[SOLVED] Projectile Motion - Ramp Angle

Good evening everyone!
1. The problem statement, all variables and given/known data
This is a classic ramp angle question. I am trying to find the angle of a ramp in order for a bus to complete the gap (Yes, this is from the film Speed, and I have searched, but haven't found a solution.)
Known:
Velocity = 31.29 m/s
Distance of gap = 15.24 m
The landing part of the road is level with the take off.

2. Relevant equations
θ = (1/2) sin^-1 (fg x d / v2).

3. The attempt at a solution
θ = (1/2) sin-1 (9.81 x 15.24 /31.292)
θ = (1/2) sin-1 (149.5044 / 979.0641)
θ = (1/2) sin-1 (0.1527)
θ = (1/2) (8.7834)
θ = 4.3917 degrees.

This answer does not quite seem right. I have gone through my calculations, but I cannot find an error. Perhaps my equation is wrong, I am not sure.

Last edited: Jan 13, 2008
2. Jan 13, 2008

### patv

I am still puzzled by this.
Could someone help me determine whether or not I am using the correct equation?

3. Jan 13, 2008

### mike115

I see that you are using the range formula. Make sure the distance is in meters. Also remember that sin(x) = sin(180 - x), so there are two solutions for the angle that yield the same range. One of the angles is 4.39 degrees while the other angle is 85.6 degrees.

4. Jan 13, 2008

### patv

Thank you. I converted the distance to meters. I doubt the angle of the ramp would have to be 85.6, so you would say 4.39 is correct?
I was expecting a number closer to around 25 degrees, but that was pure guessing.

5. Jan 13, 2008

### chickendude

The range equation is

$$R = \frac{v^2}{g}\sin{2\theta}$$

rearranging

$$\sin{2\theta} = \frac{Rg}{v^2}$$

There are two solutions for theta

$$\theta = \frac{1}{2} \sin^{-1}\frac{Rg}{v^2}$$
or
$$\theta = \frac{1}{2}[180 - \sin^{-1}\frac{Rg}{v^2}]$$

Plugging in all the numbers, you get
$$\theta = 4.39^\circ$$
or
$$\theta = 85.61^\circ$$

Any value of theta between those two will make the jump (since at either of those values, the bus barely makes it)

6. Jan 13, 2008

### patv

Thank you very much, sir.
This question had me second guessing myself because we were not given the equation, I had just found it somewhere online.
Cheers!