Just as a car tops a 42 meter high hill with a speed of 34 km/h it runs out of g

Question

Just as a car tops a 42 meter high hill with a speed of 34 km/h it runs out of gas and coasts from there, without friction or drag. How high, to the nearest meter, will the car coast up the next hill? I can't figure out what formula to use, my text book is very vague on this subject, and I have nothing to go on. I would appreciate any help! Just as a car tops a 42 meter high hill with a speed of 34 km/h it runs out of gas and coasts from there, without friction or drag. How high, to the nearest meter, will the car coast up the next hill? I can't figure out what formula to use, my text book is very vague on this subject, and I have nothing to go on. I would appreciate any help!

Explanation / Answer

   Since there is not friction and drag, net energy (k.e. + p.e.) of the car would be exactly equal to p.e. of car at the highest point on next hill.    k.e.   +   p.e.   =   p.e.next hill    (1/2) * m * v2   +   m * g * h1   =   m * g * h2    =>   0.5 * v2   +   g * h1   =   g * h2    given   v   =   34   km/h                   =   34 * 5 / 18                   =   9.44   m/s    =>   0.5 * 9.442   +   9.8 * 42   =   9.8 * h2    maximum height on next hill   h2   =   456.16 / 9.8                                                       =   46.55   m                                                       ˜   47   m    maximum height on next hill   h2   =   456.16 / 9.8                                                       =   46.55   m                                                       ˜   47   m

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