Bob shoots an arrow straight up into the air at 78.5 m/s.
1)What is the maximum height of the arrows flight?
2)How long is it in the air?
1. hmax = V^2-Vo^2)/2g.
hmax = (0-6162.25)/19.6 = 314.4 m.
2. Tr = V-Vo)/g = (0-78.5)/-9.8 = 8 s. =
Rise time.
Tf = Tr = 8 s. = Fall time.
Tr + Tf = 8 + 8 = 16 s. = Time in air.
To find the answers to these questions, we need to use the equations of motion.
1) To find the maximum height of the arrow's flight, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
where:
- vf is the final velocity (which is 0 m/s at the maximum height)
- vi is the initial velocity (in this case, upward, so it's +78.5 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2)
- d is the displacement or the maximum height we want to find.
Plugging in the values:
0 = (78.5)^2 + 2(-9.8)d
Simplifying the equation:
0 = 6162.25 - 19.6d
Solving for d:
19.6d = 6162.25
d = 6162.25 / 19.6
d ≈ 314.7 m
Therefore, the maximum height of the arrow's flight is approximately 314.7 meters.
2) To find how long the arrow is in the air, we can use the kinematic equation:
vf = vi + at
where:
- vf is the final velocity (which is 0 m/s when the arrow reaches its maximum height)
- vi is the initial velocity (in this case, upward, so it's +78.5 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time we want to find.
Plugging in the values:
0 = 78.5 - 9.8t
Solving for t:
9.8t = 78.5
t = 78.5 / 9.8
t ≈ 8.02 seconds
Therefore, the arrow is in the air for approximately 8.02 seconds.