Question

In a scene in an action movie, a stunt man
jumps from the top of one building to the
top of another building 3.8 m away. After a
running start, he leaps at an angle of 18◦ with
respect to the flat roof while traveling at a
speed of 4.2 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 1.3 m shorter than the building
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

his horizontal velocity is ... v = 4.2 m/s * cos(18º)

time to clear the gap between buildings ... t = 3.8 / v

height at end of gap ... h = -1/2 * 9.81 * t^2 + 4.2 sin(18º) t


oops ... tough luck

Answer 2

I think there are some errors above.
time in air:
hf=hi+4.2sin18*t -4.9t^2 now hf is 1.3 shorter, so
-1.3=0+4.2sin18*t -4.9t^2 Put that in quadratic form, and solve for time in air t.

Now, he has to clear the horizontal gap in that time.
distance=4.2*cos18*timeinair.
solve for distance, and see if he made it or not.

Answer 3

Vo = 4.2m/s[18o].
Yo = 4.2*sin18 = 1.30 m/s. = Ver. component.

Y^2 = Yo^2 + 2g*h1 = 0.
1.3^2 + (-19.62)h1 = 0,
h1 = 0.086 m. above 1st. bldg.
h2 = -1.3 m = 1.3 m below 1st. bldg.
Disp. = h1 - h2 = 0.086 - (-1.3) = 0.086 + 1.3 = 1.39 m.

Answer 4

Xo = 4.2*Cos18 = 3.99 m/s.
Yo + g*Tr = 0,
1.3 + (-9.81)Tr = 0,
Tr = 0.133 s. = Rise time.

4.9*Tf^2 = 1.39 m.
Tf = 0.533 = Fall time.

Range = Xo*(Tr+Tf) = 3.99(0.133 + 0.533) = 2.66 m.
Required range = 3.8 m.
So he didn't make it!!

Answer 5

there are several methods to find if the jump is successful

the 2nd part of the question specifies a solution