A parkour runner is running on top of a building when he comes to the edge; the next building is 5
meters down and 8 meters over. How fast does he need to be running to make the jump?

If we make a rather silly assumption that the runner does not leap upward somewhat as he leaves the edge of the first building, his initial vertical velocity is zero.

The time needed to drop 5 meters from vertical rest is

t = √(2s/g) = √(2(5)/9.8) = 1.01 s

so he needs a horizontal speed of at least

v = d/t = 8 / 1.01 = 7.91959...

v = 7.9 m/s. Better make it 8 m/s to be sure he clears the edge and lands on the roof.

A parkour runner is running on top of a building when he comes to the edge; the next building is 5 meters down and 8 meters over. How fast does he need to be running to make the jump?

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A parkour runner is running on top of a building when he comes to the edge; the next building is 5
meters down and 8 meters over. How fast does he need to be running to make the jump?