A Motorcyclist is trying to jump across a cliff. He will drive horizontally off

Question

A Motorcyclist is trying to jump across a cliff. He will drive horizontally off the cliff that is 65.0m high at a speed of 40 m/s. ignoring air resistance. Use energy conservation to find the speed with which the cycle strikes the ground at a height of 30.0m on the other side of the canyon. You do not need to know the how wide the canyon is.

(Vf-Vi)^2 = 2g(Hi-Hf) So Vf^2- 40m^2 Vf^2-1600m^2= 2(9.8m/s^2)(65m-40m)
Vf^2=1600m^2+2(9.8m/s^2)(65m-40m) Vf=v1600m^2+2(9.8m/s^2)(65m-40m)
Vf=v1600 m^2+19.6m/s^2(25m) Vf=v1600m^2+490m^2/s^2 Vf=v2090m^2/s^2
Vf=45.71 m/s
So, the speed at which the cycle strikes the ground is 45.71m/s, but my question pertains to the energy conservation part. Using PE=KE formulas
PElost=KEgain mgh=.5mv^2 vF=v2gh vF=v2(9.8m/s^s)(65m) vF=v1274m^2/s^2 vF=35.69 m/s
So, how can I get KE without no mass or is the first part of the problem the answer?

Explanation / Answer

Given that intial speed vi=40m/s intial height hi=65m final height hf=30m if vf is the final velocity then the change in kinetic energy is 1/2m(vf2-vi2) change in potential energy is mg(hi-hf) so at equilibrium 1/2m(vf2-vi2)=mg(hi-hf) (vf2-vi2)=2g(hi-hf) vf2=2g(hi-hf)+vi2      =2(9.8)(65-30)+402      =2286 vf =47.81m/s      =2286 vf =47.81m/s         

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