What is the Optimal Ramp Angle for a Bus Jumping a Gap?

[patv]

[SOLVED] Projectile Motion - Ramp Angle

Good evening everyone!

Homework Statement

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.

Homework Equations

θ = (1/2) sin^-1 (fg x d / v2).

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.

 

[patv]

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

 

[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.

 

[patv]

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.

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.

 

[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)

 

[patv]

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)

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!