a man is riding a motorbike at 27m/s across the top of a flat mesa. he rides str

Question

a man is riding a motorbike at 27m/s across the top of a flat mesa. he rides straight off the mesa and lands on the slopinng hill below.

a) If the difference in height between the hill and the top of the mesa is 15m, how fast is he going when he lands?

Hint: there are two ways to do this problem, with forces/kinematics and energy. Show both ways.

diagram: the motorbike is going 27m/s on a horizontal path and jumps a ditch landing on a slope shaped like a right angle triangle which is 15m lower than the path and has a angle of 32 degrees.

Explanation / Answer

By energy conservation we have .5m27^2 +mg15= .5mv^2 => 729 + 147*2 = v^2 => v= 32m/sec now this is the net velocity just before he lands on the slope. For his velocity in different direction you need to post the diagram(take a snap shot).

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