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 . What should be his minimum roof top speed , in m/s , so that he can successfully make a jump ?
To find the minimum roof top speed required for the stuntman to successfully make the jump, we can solve the problem using the principles of projectile motion.
Let's break down the problem into two components: the motion in the horizontal direction and the motion in the vertical direction.
1. Motion in the horizontal direction:
The distance between the two buildings is given as 6.2 meters. Since there are no forces acting horizontally, the horizontal component of the motion will be at a constant velocity. Therefore, there is no influence on the minimum roof top speed required for the stuntman to make the jump.
2. Motion in the vertical direction:
The vertical motion of the stuntman can be separated into two parts: ascending (from the first roof to its peak height) and descending (from the peak height to the roof of the next building).
Let's calculate the time for the ascent:
Using the equation of motion:
h = ut + (1/2)gt^2
Where:
h = vertical displacement = height difference between the two buildings (4.8 m)
u = initial vertical velocity = 0 m/s (as the stuntman starts from rest)
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
t = time taken for ascent
Rearranging the equation:
h = (1/2)gt^2
Substituting the values:
4.8 = (1/2)(9.8)t^2
Solving for t:
t^2 = (2 * 4.8) / 9.8
t^2 = 0.9796
t ≈ 0.99 s
Now, let's calculate the time for the descent:
Using the same equation of motion:
h = ut + (1/2)gt^2
Where:
h = vertical displacement = 0 m (the stuntman lands at the same height as the roof of the next building)
u = final vertical velocity (when the stuntman leaves the first roof and starts descending)
g = acceleration due to gravity = 9.8 m/s^2
t = time taken for the descent
Rearranging the equation:
0 = (1/2)gt^2
Solving for t:
t^2 = 0
t = 0 s
The total time for the entire jump is the sum of the ascent and descent times:
total time = time for ascent + time for descent
total time = t + 0
total time = t ≈ 0.99 s
The minimum roof top speed required can be obtained using the formula:
speed = distance / time
Substituting the values:
minimum roof top speed = 6.2 m / 0.99 s
minimum roof top speed ≈ 6.27 m/s
Therefore, the minimum roof top speed required for the stuntman to successfully make the jump is approximately 6.27 m/s.
Vo = Hor. speed.
R = 6.2 m. = hor. range.
0.5g*T^2 = 4.8 m.
4.9*T^2 = 4.8,
T = 0.99 s.
R = Vo * T = 6.2 m.
Vo * 0.99 = 6.2,
Vo = ?