NO!!!!!!!

Why would there be an acceleration in the horizontal direction?

Define an origin at the point of take-off, and define the +x direction to be the direction the motorcycle is traveling in before the jump, and the +y direction to be upward. Then we know that at the point of takeoff:

x0 = 0 m

y0 = 0 m

v0x = ?

v0y = 0 m/s

ax = 0 m/s^2

ay = -g = -9.8 m/s^2

At the end of the jump:

x = 5.1 m

y = -2 m

t = ?

So we want to know v0x. The only equation we have that we can work with here is:

x = x0 + v0x*t

5.1 = v0x*t

We know neither v0x nor t. So we turn to the y information. We have 4 equations at our disposal:

y = y0 + v0y*t + (1/2)ay*t^2

y = y0 + (1/2)(v0y + vy)t

vy = v0y + ay*t

vy^2 = v0y^2 + 2ay*(y - y0)

We need a value for t and the only equation up here with only one unknown is the top one. So:

y = y0 + v0y*t + (1/2)ay*t^2

-2 = (1/2)(-9.8)t^2

t^2 = 4/9.8 = 0.40816

t = 0.63888 s

Thus

5.1 = v0x*[0.63888]

v0x = 5.1/0.63888 = 7.9828 m/s

Or 8.0 m/s to 2 sig. digs.

-Dan