Impact force on a bike off a 6 foot drop?

aliikane

Here is a question for the physics experts. Some people been debating about the impact force on a bike of a bike rider riding off a 6 foot drop to a flat landing. Mainly the debate is about if speed would lessen the impact force on the bike because of the impact angle.

So, if a bike rider was to ride off a 6 foot drop, would it lessen the impact force on the bike jumping off the 6 foot drop with faster speed rather than slower speed and just drop off? In both examples the bike rider is landing on both tires exactly at the same time.

Thanks in advance for the reply.

berkeman

Welcome to the PF.

I don't think the horizontal speed will make a difference. A flat-lander is a flat-lander, and it's going to hurt some, depending on your bike's suspension.

Gnosis

Berkeman is correct, the horizontal velocity will not lessen the vertical impact force.

Let’s presume the bike begins its drop from a 1.8288 meter (6 foot) height at a horizontal velocity of 13.4 m/s (approx. 30 MPH) and at the same time, a rock with the same mass as the bike is released to free-fall directly to the ground below from the same 1.8288 meters. Both the bike and the rock will simultaneously make contact with the ground, as they are being accelerated by the same gravitational rate of 9.8 m/s^2 therefore, both acquired precisely the same vertical velocity. Since both have the same vertical descent velocity, the bike’s horizontal velocity couldn’t possibly have reduced the vertical impact force that it will experience, as that vertical velocity of the bike is abruptly made zero upon impacting the unforgiving ground no differently than that of the rock from the same height. In the vertical direction, both deliver precisely the same impact force per their identical mass, as per the kinetic energy equation:

E = ½ * m * (v)^2

Both objects would free-fall from a height (s) of 1.8288 meters in just .61 seconds per an earthly gravitational acceleration (a) of 9.8 m/s^2 per the following kinematics equation:

t = sqrt(2s / a)

sqrt(2 * 1.8288 meters / 9.8 m/s^2) = .61 seconds

The bike would simply land further away than the rock per the product of the bike’s 13.4 m/s horizontal velocity and its free-fall time of .61 seconds. Since no velocity is lost in the horizontal direction, it becomes evident that the energy associated with the horizontal velocity couldn’t possibly have been applied to reduce the vertical velocity that had been acquired during the bike’s free-fall, for had work been performed, the horizontal velocity would have been diminished (by more than the trivial air resistance).

I hope you found this helpful.