Projectile Motion with different starting and ending heights

[kayman121]

Homework Statement

A motorcyclist is going off an incline 53 degrees above the horizontal. He wants to land on a ledge 40 meters from the ledge he is launching off of, 15 meters below the edge of the ramp he is laughing off of. Δx = 40m. Δy = -15m. What is the minimum initial velocity he needs to go off the ramp at to barely make the jump?

Homework Equations

Vx = Vocos53
Ag = 9.8 m/s^2
Voy = Vosin53
x = Vx(t)
t = ?
y = yo + Voy(t)-1/2(a)(t)^2

The Attempt at a Solution

If this was a horizontal launch it'd be very simple, but I have no idea what to do with launching off an incline. I tried:

t = square root (2d/a) = square root ((2*15)/9.8) = 1.75 sec

x = Vx(t) therefore 40 = Vx(1.75) ; Vx = 22.86 m/s

Then; Vo = Vx/cos53 = Vo = 22.66/cos53 = 37.99 m/s. The book says this is wrong and I have no idea what to do.

 

[ideasrule]

t = square root (2d/a) = square root ((2*15)/9.8) = 1.75 sec

t does not equal the square root of 2d/a because the initial vertical speed is not 0. Do you know which equation 2d/a came from? Use that equation; you now have an equation relating vertical displacement (15 m), time, and initial speed. There's one other equation that you need, and it's relating to horizontal speed.

 

[kayman121]

t does not equal the square root of 2d/a because the initial vertical speed is not 0. Do you know which equation 2d/a came from? Use that equation; you now have an equation relating vertical displacement (15 m), time, and initial speed. There's one other equation that you need, and it's relating to horizontal speed.

Unfortunately, that leaves 2 unknowns in 1 equation. The unknowns being initial speed and time.

 

[ideasrule]

I said that you need one other equation--namely, the one relating to horizontal speed.