In a scene in an action movie, a stunt man jumps from the top of one building to the top of another building 5.0 m away. After a running start, he leaps at an angle of 10◦ with respect to the flat roof while traveling at a speed of 5.6 m/s.
The acceleration of gravity is 9.81 m/s2 .
To determine if he will make it to the other roof, which is 3.0 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 = 5.6m/s[10o].
Xo = 5.6*Cos10 = 5.51 m/s.
Yo = 5.6*sin10 = 0.97 m/s.
Y^2 = Yo^2 + 2g*h = 0.
0.97^2 - 19.6h = 0, h = 0.048 m.
Y = Yo + g*Tr = 0.
0.97 - 9.8Tr = 0, Tr = 0.099 s. = Rise time.
0.5g*Tf^2 = (0.048+3).
4.9Tf^2 = 3.048, Tf = 0,789 s. = Fall time.
Dx = Xo*(Tr+Tf) = 5.51m/s * (.099+0.789)s. = 4.89 m.
But the distance between buildings is 5 m.
To determine the stuntman's vertical displacement upon reaching the front edge of the lower building with respect to the taller building, we can break down the motion into horizontal and vertical components.
First, let's find the time it takes for the stuntman to reach the other building. We will assume air resistance is negligible.
The horizontal distance between the two buildings is 5.0 m, and his horizontal speed is given as 5.6 m/s. We can use the formula:
distance = speed × time
Rearranging the formula, we have:
time = distance / speed
Substituting the values, we get:
time = 5.0 m / 5.6 m/s
time ≈ 0.893 s
Now let's find the vertical displacement of the stuntman using the vertical motion equation:
displacement = initial velocity × time + (1/2) × acceleration × time^2
The initial vertical velocity is determined by the vertical component of his initial velocity. We can calculate it using trigonometry:
initial vertical velocity = initial velocity × sin(angle)
Substituting the values, we have:
initial vertical velocity = 5.6 m/s × sin(10°)
initial vertical velocity ≈ 0.970 m/s
Since the stuntman jumps from the top of the taller building, we can assume his initial vertical displacement is 0.
Now, substituting the values into the vertical motion equation, we have:
displacement = (0.970 m/s) × (0.893 s) + (1/2) × (9.81 m/s^2) × (0.893 s)^2
Calculating this expression, we find:
displacement ≈ 0.409 m
Therefore, his vertical displacement upon reaching the front edge of the lower building with respect to the taller building is approximately 0.409 meters.