physics
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
◦
with
respect to the flat roof while traveling at a
speed of 5.8 m/s.
The acceleration of gravity is 9
.
81 m
/
s
2
.
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
building.
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 = (YYo)/g = (01.70)/9.8 = 0.173 s.
= Rise time.
h=(Y^2Yo^2)/2g = (01.7^2)/19.6=0.147 m.
h = 1.4 + 0.147 = 1.55 m Above shorter
bldg.
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.
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