In a scene in an action movie, a stunt man jumps from the top of one building to the top of another building 5.0 m away. After a running start, he leaps at an angle of 10◦ with respect to the flat roof while traveling at a speed of 5.6 m/s.

Question

The acceleration of gravity is 9.81 m/s2 .
To determine if he will make it to the other roof, which is 3.0 m shorter than the build- ing from which he jumps, find his vertical displacement upon reaching the front edge of the lower building with respect to the taller building.
Answer in units of m.

Answer 1

Vo = 5.6m/s[10o].
Xo = 5.6*Cos10 = 5.51 m/s.
Yo = 5.6*sin10 = 0.97 m/s.

Y^2 = Yo^2 + 2g*h = 0.
0.97^2 - 19.6h = 0, h = 0.048 m.

Y = Yo + g*Tr = 0.
0.97 - 9.8Tr = 0, Tr = 0.099 s. = Rise time.

0.5g*Tf^2 = (0.048+3).
4.9Tf^2 = 3.048, Tf = 0,789 s. = Fall time.

Dx = Xo*(Tr+Tf) = 5.51m/s * (.099+0.789)s. = 4.89 m.
But the distance between buildings is 5 m.