posted by Anonymous .
In a scene in an action movie, a stunt man
jumps from the top of one building to the
top of another building 4.9 m away. After a
running start, he leaps at an angle of 17
respect to the flat roof while traveling at a
speed of 5.8 m/s.
The acceleration of gravity is 9
To determine if he will make it to the other
roof, which is 1.4 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
Answer in units of m
Vo = 5.8m/s[17o]
Xo = 5.8*cos17 = 5.55 m/s.
Yo = 5.8*sin17 = 1.70 m/s.
Y = Yo + g*t = 0 @ max. ht.
Tr = (Y-Yo)/g = (0-1.70)/-9.8 = 0.173 s.
= Rise time.
h=(Y^2-Yo^2)/2g = (0-1.7^2)/-19.6=0.147 m.
h = 1.4 + 0.147 = 1.55 m Above shorter
h = 0.5g*t^2 = 1.55 m.
4.9*t^2 = 1.55
t^2 = 0.316
Tf = 0.562 s. = Fall time or time to fall to roof of shorter bldg.
Dx = Xo * (Tr+Tf)
Dx = 5.55m/s * (0.173+0.562) = 4.08 m.
= His Hor. distance from bldg. #1.
Therefore, he will fall short of bldg. #2:
4.9 - 4.08 = 0.82 m. Short.