Question

For women's volleyball the top of the net is 2.24 m above the floor and the court measures 9.00 m by 9.00 m on each side of the net. Using a jump serve, a player strikes the ball at a point that is 2.91 m above the floor and a horizontal distance of 8.02 m from the net. If the initial velocity of the ball is horizontal, (a) what minimum magnitude must it have if the ball is to clear the net and (b) what maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net?

Answer 1

Answer:

a) The minimum magnitude of the initial velocity the ball must have to clear the net is 21.7 m/s.

b) The maximum magnitude of the initial velocity the ball must have to strike the floor inside the back line is 22.1 m/s.

Explanation:

Please see the figure for a description of the problem.

The equation for the vector position is:

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

Where:

r = position vector

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

g = acceleration due to gravity

v0y = initial vertical velocity  

a)Seeing the figure, we can notice that the x-component and y-component of the vector "r" is 8.02 m and 2.24 m respectively.

Then:

r = (8.02 m, 2.24 m)

with this data, we can obtain the time at which the ball passes the net:

Using the y-component equation of the vector "r"

y = y0 + v0y * t + 1/2 * g * t² = 2.24 m        v0y = 0 and y0 = 2.91 m

2.91 m - 1/2 (9.8 m/s²) * t² = 2.24 m

- 1/2 (9.8 m/s²) * t² = 2.24 m - 2.91 m

t² = - 0.67 m/ -4.9 m/s²

t = 0.37 s

With this time, we can obtain the initial velocity from the equation for the position of the x-component of the vector "r".

x = x0 + v0x * t = 8.02 m

Since the origin of the reference system is located at the point at which the ball is hit but on the ground, x0 = 0. Then:

v0x * t = 8.02 m

v0x = 8.02 m / t

v0x = 8.02 m / 0.37 s = 21.7 m/s

The minimum magnitude of the initial velocity the ball must have to clear the net is 21.7 m/s.

b) To solve this, we can proceed as in the previous question but using the vector "r final". From the figure, notice that the vector "r final" has the following x- and y-components:

r final = (8.02 m + 9.00 m, 0 m) = (17.02 m, 0 m)

Again, we use the equation of the y-component of the vector position to obtain the time and with that time, we can calculate v0x.

y = y0 + v0y * t + 1/2 * g * t² = 0

2.91 m - 1/2 (9.8 m/s² ) * t² = 0

t² = -2.91 m / (-4.9 m/s²)

t = 0.77 s

Now we replace this time in the equation of the x-component of the vector position:

x = x0 + v0x * t = 17.02 m

v0x * 0.77 s = 17.02 m

v0x = 17.02 m / 0.77 s = 22.1 m/s

The maximum magnitude of the initial velocity the ball must have to strike the floor inside the back line is 22.1 m/s.