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
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
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.
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.
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!!
there are several methods to find if the jump is successful
the 2nd part of the question specifies a solution
To solve this problem, we need to break down the motion of the stuntman into horizontal and vertical components.
First, let's find the time it takes for the stuntman to reach the top of the other building. We can use the horizontal component of the motion to determine this.
The horizontal component of the velocity remains constant throughout the motion. So the horizontal velocity (Vx) can be calculated using the initial velocity (4.2 m/s) and the angle (18 degrees):
Vx = V * cos(angle)
Vx = 4.2 m/s * cos(18 degrees)
Next, we can calculate the time it takes for him to travel the 3.8 m horizontal distance using the formula:
Distance = Velocity * Time
3.8 m = Vx * Time
Solving for Time:
Time = 3.8 m / Vx
Now, let's find the vertical displacement (height) of the stuntman upon reaching the front edge of the lower building with respect to the taller building.
The vertical displacement can be found using the equation:
Vertical displacement = Vy * Time + (0.5 * acceleration * Time^2)
where Vy is the initial vertical velocity and acceleration is the acceleration due to gravity (-9.81 m/s^2).
To find the initial vertical velocity, we need to find the vertical component of the initial velocity using the same equation:
Vy = V * sin(angle)
Vy = 4.2 m/s * sin(18 degrees)
Plugging in the values, we can now calculate the vertical displacement:
Vertical displacement = (4.2 m/s * sin(18 degrees)) * Time + (0.5 * -9.81 m/s^2 * Time^2)
Substituting the value of Time we derived earlier:
Vertical displacement = (4.2 m/s * sin(18 degrees)) * (3.8 m / Vx) + (0.5 * -9.81 m/s^2 * (3.8 m / Vx)^2)
Calculating this expression will give us the vertical displacement of the stuntman upon reaching the front edge of the lower building with respect to the taller building.
Let me calculate it for you.