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.

Answer in units of m

- 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.

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