A volleyball is deflected vertically by a player in a game housed in a high school gymnasium where the ceiling clearance is 18m. If the initial velocity of the ball is 21.5 m/s, will the ball contact the ceiling?
To determine if the ball will contact the ceiling, we need to calculate the maximum height it reaches.
We can use the kinematic equation to calculate the maximum height (h) reached by the ball:
vf^2 = vi^2 + 2 * a * h
Where:
- vf is the final velocity (0 m/s at the maximum height)
- vi is the initial velocity (21.5 m/s)
- a is the acceleration (due to gravity, -9.8 m/s^2)
- h is the maximum height reached by the ball
Rearranging the equation to solve for h, we get:
h = (vf^2 - vi^2) / (2 * a)
Let's substitute the given values into the equation:
h = (0^2 - 21.5^2) / (2 * -9.8)
h = (-462.25) / -19.6
h ≈ 23.56 meters
Therefore, the ball reaches a maximum height of approximately 23.56 meters. Since this is higher than the ceiling clearance of 18 meters, the ball will indeed contact the ceiling.
To determine if the volleyball will contact the ceiling, we need to calculate the maximum height the ball reaches. If the maximum height is less than the ceiling clearance of 18m, then the ball will not contact the ceiling.
To find the maximum height, we can use the following kinematic equation:
v_f^2 = v_i^2 + 2aΔy
Where:
- v_f is the final velocity (0 m/s at the maximum height)
- v_i is the initial velocity (21.5 m/s)
- a is the acceleration (assumed to be -9.8 m/s^2 due to gravity)
- Δy is the change in height (the maximum height we want to find)
Rearranging the equation to solve for Δy, we have:
Δy = (v_f^2 - v_i^2) / (2a)
Substituting the values, we get:
Δy = (0 - (21.5 m/s)^2) / (2 * -9.8 m/s^2)
= -465.25 m^2/s^2 / -19.6 m/s^2
≈ 23.74 m
Since the result is positive, the ball will reach a maximum height of 23.74 meters. As this height is less than the ceiling clearance of 18 meters, the ball will not contact the ceiling.
d=vot+1/2at^2
vf=vo+at
V=d/t
Use these equations to solve the problem.