How fast must the motorcycle leave the cliff top to land on a level ground below?

Nicholas: How fast must the motorcycle leave the cliff top to land on a level ground below?
A movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast must the motorcycle leave the cliff top to land on a level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.

Answers and Views:

Answer by Debbie
From the height you can determine the time it takes for the cycle to hit the ground from y=1/2gt^2. So t = sqrt(2y/g) = sqrt (2*50/9.8) = 3.19s. Since there is no acceleration in the x direction we use v=d/t so v = 90m/3.19s = 28.2m/s

Answer by Jorge
This sounds like a homework problem. lol. It isn’t that hard. I am not going to give you an answer but will tell you how to solve it and you can take it from there. The frist thing you have to understand is that regardless of the speed horizontal to the ground the bike will only have a given amount of time in the air. To find this time you must calculate the time it takes the bike to fall the distance of 50m under the force of gravity. Then this time is the total air time. With the understanding that the bike is at constant velocity once it is airborne your next task is to calculate the velocity required to travel 90m in the time you found before. Remember v=d/t and as for the time required for it to fall i would look up in you book (or on google) the free fall formula but i think it is .5(a)(t)^2=distance

Teach a man to fish.

Answer by Dude
just fast enough to make it over the cliff –gravity will do the rest