In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building....?

Question

 In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building. After a running start, he leaps at a velocity of 5.0 m/s at an angle of 15 degrees with respect to the flat roof. If the other roof is 2.5 m below the building he jumps from, how far away can the other building be so he safely makes the jump?

Update:

Physics question 

Answer 1

Favorite Answer

h-ho = Vo*sin 15°*t-g/2*t^2

-2.5 = 5*0.259*t-4.9*t^2 

-2.5-1.3t+4.9t^2 = 0

t = (1.3+√ 1.3^2+2.5*4*4.9)/9.8 = 0.86 sec 

d < Vo*t 

d < 4.3 m 

Answer 2

Use the trajectory equation so you don't need to find the time:

y = h + x*tanΘ - gx² / (2v²cos²Θ)

Dropping units for ease,

0 = 2.5 + x*tan15º - 9.8x²/(2*5.0²cos²Θ) = 2.5 + 0.268x - 0.21x²

quadratic with roots at x = -2.9 m ← ignore

and x = 4.1 m ◄

Answer 3

i just liked the movie                 

Answer 4

He will be hooked up to a wire harness, so the other building could theoretically be on the other side of town.