A stuntman plans to run across a roof top and then horizontally off it to a land on the roof of the next building . The roof of the next building is 4.8 m below the first one and 6.2 m away from it . Can he make the jump if his maximum rooftop speed is 4.5 m/s?
Vo = 4.5 m/s, horizontally.
0.5g*t^2 = 4.8 m.
4.9*t^2 = 4.8,
t = 0.99 s. = Fall time.
Range = Vo * t = 4.5 * 0.99 = 4.45 m.
Required range = 6.2 m. No, he will not make the jump!
To determine if the stuntman can make the jump, we need to find out if he can cover the horizontal distance and vertical drop within his maximum rooftop speed.
Let's first calculate the time it takes for the stuntman to land on the next building's roof.
The horizontal distance is 6.2 m, and the maximum speed is 4.5 m/s. We can use the formula: time = distance / speed.
time = 6.2 m / 4.5 m/s = 1.3778 s (rounded to four decimal places)
Now, let's calculate the distance the stuntman drops vertically during this time.
The vertical drop is 4.8 m. We can use the formula of vertical displacement under constant acceleration: distance = initial velocity × time + (1/2) × acceleration × time^2.
Since the stuntman starts from rest vertically, the initial velocity is 0 m/s, and the acceleration is due to gravity: 9.8 m/s².
distance = 0 × 1.3778 s + (1/2) × 9.8 m/s² × (1.3778 s)^2
distance = 8.0346 m (rounded to four decimal places)
The stuntman lands 8.0346 m below the initial roof level, which is more than the 4.8 m drop to the next building's roof.
Therefore, the stuntman can successfully make the jump with his maximum rooftop speed of 4.5 m/s.
To determine if the stuntman can make the jump, we need to consider his horizontal and vertical velocities.
Let's break down the problem into two components:
1. Vertical Component:
- The height difference between the two rooftops is 4.8 m.
- We can calculate the time it takes for the stuntman to fall from the first rooftop to the second rooftop using the equation:
h = (1/2) * g * t^2
where h is the height difference, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
- Rearranging the equation to solve for t, we have:
t = sqrt((2 * h) / g)
- Substituting the given values, we get:
t = sqrt((2 * 4.8) / 9.8)
- Calculating the value of t, we find t ≈ 0.98 seconds.
2. Horizontal Component:
- The horizontal distance between the two buildings is 6.2 m.
- We can calculate the time it takes for the stuntman to cover this horizontal distance using the equation:
d = v * t
where d is the distance, v is the velocity, and t is the time.
- Rearranging the equation to solve for t, we have:
t = d / v
- Substituting the given values, we get:
t = 6.2 / 4.5
- Calculating the value of t, we find t ≈ 1.38 seconds.
Now, we need to compare the times for the vertical and horizontal components. If the time taken for the vertical component (falling) is less than or equal to the time taken for the horizontal component (running), then the stuntman can make the jump.
In this case, the vertical time (0.98 seconds) is less than the horizontal time (1.38 seconds), so the stuntman can make the jump.