Thread: Horizontal Speed - Physics

A stuntman on a motorcycle jumps a river which is 5.1m wide.
He lands on the edge of the far bank, which is 2.0m lower than the bank from which he takes off.

His minimum horizontal speed at take of is?

The answer = 8.0ms^-1

Could someone tell me which formula to use?

Thanks

Nevermind, just realised the equation was 's = ut + 0.5at^2' a = 9.8


(remove this thread if needed)

Originally Posted by r_maths

A stuntman on a motorcycle jumps a river which is 5.1m wide.
He lands on the edge of the far bank, which is 2.0m lower than the bank from which he takes off.

His minimum horizontal speed at take of is?

The answer = 8.0ms^-1

Could someone tell me which formula to use?

Thanks

Nevermind, just realised the equation was 's = ut + 0.5at^2' a = 9.8


(remove this thread if needed)

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