In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building....?
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?
- oubaasLv 71 month agoFavorite 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
- NCSLv 71 month ago
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 ◄
- Anonymous1 month ago
i just liked the movie
- Little PrincessLv 71 month ago
He will be hooked up to a wire harness, so the other building could theoretically be on the other side of town.