Wednesday
December 7, 2016

# Homework Help: physics

Posted by Anonymous on Thursday, November 7, 2013 at 9:18pm.

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.
• physics - Henry , Friday, November 8, 2013 at 6:31pm

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
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.