As a movie stunt coordinator, you need to be sure a stunt will be safe before it is performed. If a stunt- woman is to run horizontally off the roof of a three-story building and land on a foam pad 7.35 m away from the base of the building, how fast should she run as she leaves the roof? (Each story is 4.00 m.)
h = ½ g t² ... t = √(2 * 12.0 / g)
v = d / t = 7.35 m / t
To determine the required speed for the stuntwoman to run off the roof, we can use the kinematic equation that relates distance, initial velocity, time, and acceleration. In this case, we can assume that the acceleration is due to gravity and is approximately 9.8 m/s².
The horizontal distance traveled by the stuntwoman is 7.35 m. This distance is equal to the horizontal component of her velocity (which remains constant) multiplied by the time of flight.
In this scenario, the height of the three-story building is 3 stories × 4.00 m/story = 12.00 m.
To calculate the time of flight, we can use the equation:
h = 1/2 * g * t^2,
where h is the height, g is the acceleration due to gravity, and t is the time of flight.
By re-arranging the equation, we get:
t = sqrt(2h / g).
Let's plug in the values:
t = sqrt(2 * 12.00 m / 9.8 m/s²)
= sqrt(2.44898 s²)
= 1.5635 s.
Now, let's calculate the initial velocity (v) using the equation:
d = vt,
where d is the horizontal distance traveled, v is the initial velocity, and t is the time of flight.
Plugging in the values:
7.35 m = v * 1.5635 s.
Solving for v:
v = 7.35 m / 1.5635 s
= 4.6952 m/s.
Therefore, the stuntwoman should run off the roof with a speed of approximately 4.70 m/s.
To determine the speed the stuntwoman should run as she leaves the roof, we can use the principles of projectile motion. Here's how you can calculate it:
1. First, convert the height of the three-story building into meters. Since each story is 4.00 m, the total height of the building will be 4.00 m × 3 = 12.00 m.
2. Next, let's break down the horizontal and vertical components of the stuntwoman's motion. The horizontal distance she needs to cover is given as 7.35 m, and the vertical height is 12.00 m.
3. The horizontal component of motion (distance) is constant and does not depend on the vertical component. Therefore, the horizontal velocity (Vx) remains the same throughout. We can find the time it takes for the stuntwoman to cover this horizontal distance.
Using the formula: distance = velocity × time,
We have 7.35 m = Vx × t.
4. Solving for time (t), we get: t = 7.35 m / Vx.
5. Now, let's consider the vertical component of motion. Using the formula for vertical displacement in free-fall motion:
Δy = Vyi × t + (1/2) × g × t²,
where Δy is the vertical displacement, Vyi is the initial vertical velocity (which is 0 since she starts from rest), g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time calculated in step 4.
Plug in the values and simplify the equation to solve for Vyi:
12.00 m = 0 × t + (1/2) × 9.8 m/s² × t².
12.00 m = 4.9 m/s² × t².
6. Rearranging the equation, we have: t² = 12.00 m / (4.9 m/s²) = 2.449 sec². Taking the square root of both sides, we find: t ≈ 1.565 sec.
7. Revisit the horizontal component equation from step 3: t = 7.35 m / Vx. Substitute the calculated value of t (1.565 sec) into this equation and solve it for Vx:
1.565 sec = 7.35 m / Vx.
Vx = 7.35 m / 1.565 sec.
8. Finally, divide the horizontal distance 7.35 m by the time 1.565 sec to find the speed (Vx) the stuntwoman should run on the roof:
Vx ≈ 4.69 m/s.
So, the stuntwoman should run approximately 4.69 meters per second (m/s) as she leaves the roof to reach the foam pad 7.35 m away.