Suppose the height of the net is 0.800 m , and the court's border is 11.2 m from the bottom of the net. Use g =10 m/s2 and find the maximum speed of the horizontally moving ball clearing the net..
Ok, it is a projectile problem.
a. find the initial vertical speed. The at the mazimum height, its vertical speed is zero.
vf^2=vi^2+2ah in the vertical
0=vi^2)+2 (-10)h
vi=sqrt(20*.8) =sqrt10m/s=4m/s
now, that is the initial vertical speed.
ok, next is the time it takes the ball to reach the maximum point.
In the vertical=
vf=vi-gt
0=4m/s-10t or t=.4sec
so the horizontal velocity is now 11.2/4 m/s That is the max horizonall ball.
The initial velocity is now
vi=sqrt(4^2+(11.2/4)^2 )
To find the maximum speed of the horizontally moving ball clearing the net, we need to use the concept of projectile motion.
First, let's identify the given information:
- The height of the net (h) = 0.800 m
- The distance from the bottom of the net to the court's border (d) = 11.2 m
- Acceleration due to gravity (g) = 10 m/s^2
Now, to find the maximum speed of the horizontally moving ball, we can use the following steps:
Step 1: Determine the time taken by the ball to reach the net height.
We can use the equation of motion in the vertical direction:
h = (1/2)gt^2
Rearranging the equation, we get:
t^2 = (2h)/g
Substituting the given values, we have:
t^2 = (2 * 0.800) / 10
Simplifying, we get:
t^2 = 0.160
t ≈ 0.400 s (taking the positive square root)
Step 2: Calculate the horizontal distance covered by the ball during this time.
Since the ball is moving horizontally, its horizontal velocity remains constant throughout.
The horizontal distance covered (d) is given by:
d = v*t
Where v is the horizontal velocity and t is the time taken.
Substituting the values, we have:
11.2 = v * 0.400
Simplifying, we find:
v = 11.2 / 0.400
v ≈ 28 m/s
Therefore, the maximum speed of the horizontally moving ball clearing the net is approximately 28 m/s.