In a chase scene, a movie stuntman runs horizontally off the flat roof of one building and lands on another roof 2.0m lower.
If the gap between the buildings is 4.4m wide, how fast must he run to cross the gap?
Express your answer to two significant figures and include the appropriate units.
h = 0.5g*t^2 = 2 m.
4.9t^2 = 2
t^2 = 0.408
Tf = 0.639 s. = Fall time.
Xo*Tf = 4.4 m
Xo * 0.639 = 4.4
Xo = 6.89 m/s.
To find the speed at which the stuntman must run to cross the gap, we can use the equation of motion:
υ^2 = u^2 + 2as
Where:
υ = final velocity
u = initial velocity
a = acceleration
s = displacement
In this case, the final velocity is zero since the stuntman lands on the other roof. The initial velocity is what we want to find, and the displacement is the gap between the buildings.
Since the stuntman runs horizontally, there is no vertical acceleration acting on him. Therefore, the only acceleration acting on the stuntman is due to gravity, which can be approximated as 9.8 m/s^2.
Applying the equation of motion:
0^2 = u^2 + 2(9.8)(-2.0)
Simplifying:
0 = u^2 - 39.2
Rearranging the equation:
u^2 = 39.2
Taking the square root of both sides:
u ≈ √39.2
u ≈ 6.3 m/s
Therefore, the stuntman must run at approximately 6.3 m/s to cross the 4.4m wide gap between the buildings.
To find the speed at which the stuntman must run to cross the gap, we can use the principles of projectile motion. Let's break down the problem into different parts:
1. Vertical Motion: The stuntman jumps downward from the first roof to the lower second roof, with a vertical displacement of 2.0m. We can use the equation for vertical displacement in freefall:
Δy = Vyi * t + (1/2) * g * t^2
Here, Δy represents the vertical displacement, Vyi is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Since the stuntman lands on the second roof, Δy = -2.0m, Vyi = 0 (as he starts from rest vertically), and g = 9.8 m/s^2, we can simplify the equation to:
-2.0 = (1/2) * 9.8 * t^2
-4.0 = 9.8 * t^2
Solving for t, we find:
t^2 = -4.0 / 9.8
t ≈ 0.64 seconds
2. Horizontal Motion: The stuntman must cross a horizontal distance of 4.4m between the roofs. The horizontal motion is uniform (constant velocity) since there is no horizontal acceleration involved. We can use the equation for horizontal displacement:
Δx = Vx * t
Here, Δx represents the horizontal displacement, Vx is the horizontal velocity, and t is the time of flight, which we found to be approximately 0.64 seconds.
Since Δx = 4.4m and t = 0.64 seconds, we can solve for Vx:
4.4 = Vx * 0.64
Solving for Vx, we find:
Vx = 4.4 / 0.64 ≈ 6.88 m/s
Rounding to two significant figures, the stuntman must run at a speed of approximately 6.9 m/s to cross the gap between the buildings.