Question

Suppose you want to leap from the top of a building to the top of an adjacent building of the same height, across a gap of 3.87 m. With what minimum initial velocity would you have to jump?

magnitude
direction	above the horizontal

Answer 1

for maximum horizontal distance angle should be 45 degrees

so range x is given by

X = v²/g for projection angle of 45 degrees

X = 3.87 m

g = 9.8 m/s²

that gives v = 6.16m/s

so the answer is

magnitude: 6.16m/s

directionabove the horizontal: 45 degrees

Answer 2

Range = vo^2*sin(2*theta)/g

R is max for max sin (2*theta)

max of sin (2*theta) = 1

3.87 = Vo^2 (1)/9.8

Vo = 6.16 m/s

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sin (2*theta) = 1

theta = 45

diretion 45 defgrees above horizantal