Projectile Motion: Finding the correct angle for a launch

[Simon777]

Homework Statement

A Hollywood daredevil plans to jump the canyon shown in the figure 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 69.4 m across. If he desires a 3.0-second flight time, what is the correct angle for his launch ramp (deg)?

Homework Equations

vx = vxo

x − xo = (vox)t

(vy) = (vyo) − gt

y − yo = (voy)t −(1/2)g t^2

Range =((vo^2)/g) sin(2θ)

The Attempt at a Solution

x − xo = (vox)t
69.4m = (vox)(3s)
23.13 m/s = vox

Range =((vo^2)/g) sin(2θ)
69.4m ={ (23.13^2 m/s)/9.81 m/s^2 } sin(2θ)
θ = error

 

[ehild]

Range =((vo^2)/g) sin(2θ)

This formula is valid when the initial and final heights are the same. It is not true now, as there is a 15 m drop.

ehild

 

[Simon777]

This formula is valid when the initial and final heights are the same. It is not true now, as there is a 15 m drop.

ehild

Didn't realize this formula had that limitation. How would you find the correct angle without it? I have no other formulas involving theta.

 

[ehild]

You have the formulae

x=v_0xt
y=y₀+v_0yt-g/2 t².

You know t, the horizontal distance and the height from where the motorcycle jumps (15 m) with respect to the final height (zero). Can you get v_0y?
If you have both the x and y components of the velocity, can you find its angle with the positive x axis?

ehild

 

[Simon777]

You have the formulae

x=v_0xt
y=y₀+v_0yt-g/2 t².

You know t, the horizontal distance and the height from where the motorcycle jumps (15 m) with respect to the final height (zero). Can you get v_0y?
If you have both the x and y components of the velocity, can you find its angle with the positive x axis?

ehild

I went that route at first and ended up with a wrong answer. I've checked my calculations 5 times and keep coming up with the same wrong angle so if you wouldn't mind double checking me, I'd appreciate it.

y − yo = (voy)t −(1/2)g t^2
15m = (voy) (3s) - (1/2) (9.81m/s^2)(3s)^2
59.145 = (voy) (3s)
voy = 19.7 m/s

x − xo = (vox)t
69.4m = (vox)(3s)
23.13 m/s = vox

To find the angle I used:
tan^-1 of (19.715/23.13) = 40.44 degrees

 

[ehild]

y − yo = (voy)t −(1/2)g t^2
15m = (voy) (3s) - (1/2) (9.81m/s^2)(3s)^2

The problem says that the motorcycle drops 15 m. Your formula means that it raises as your final position is 15 m higher than the initial one. Correct it.

ehild

 

[Simon777]

The problem says that the motorcycle drops 15 m. Your formula means that it raises as your final position is 15 m higher than the initial one. Correct it.

ehild

I finally see where I went wrong. I chose down as the positive direction for y to not have to deal with negative numbers, hence the positive acceleration and positive 15m.

Where this went wrong is that the formula already factors a negative downward acceleration into it with the -(1/2)gt^2 so the downward direction has to be negative.

Thank you so much for helping me realize this.

I ended up with the right answer of 22.8 degrees.