motorcycle jump

Question

 In an action movie, the heroine is supposed to jump with a motorcycle from the roof of one skyscraper to another one. The roof of the second skyscraper is h=8 meters lower, and the gap between the buildings is d=14 meters wide. Both roofs are horizontal and flat. Neglecting air resistance, what minimum initial speed would she need to make the jump

Answer 1

To do this problem, you need to split it up into two separate problems: a) what is happening in the y-direction, and b) what is happening in the x-direction.

Y-Direction:

Since the motorcycle/heroine will be traveling initially straight across the top of the 1st roof, they will leave the edge with a velocity of zero in the y-direction (all their velocity is in the x-direction). Having a velocity of zero in the y-direction is the same as free-falling. We can use:

x = v₀t + 1/2at² where x = the 8 meter height change, v₀ is the initial velocity in the y-direction (which is zero) and "a" id=s the acceleration due to gravity, aka "g" which is 9.81 m/s²

So: 8 = (0)t + 1/2(9.81)t²

8 = 4.905t²

t = √(8/4.905) = 1.2771 seconds.

X-Direction:

During the 1.2771 seconds the motorcycle/heroine must travel the 14 meters to the other building. Therefore their minimum speed would be 14/1.2771 = 10.96 m/s which I would round to 11 m/s