Question

A few years ago, Serena Williams dived to hit a tennis ball right after it bounced off the ground. The ball bounced on the ground 11.7 m from the net, and after Serena hit the ball it flew over the 0.950 m high net and bounced in her opponent's court about 1.21 s after she hit it. If there had been no gravity, the ball would have been 2.63 m higher than the net when it crossed over. How fast was the ball moving when it left Serena's racket?

Answer 1

Answer:

The initial velocity of the ball was 20 m/s

Explanation:

Please, see the figure for a description of the problem.

The initial velocity vector can be written as follows:

v0 = (v0x, v0y)

where:

v0 = initial velocity

v0x = horizontal component of the initial velocity

v0y = vertical component of the initial velocity

The position and velocity of the ball at time "t" are described by the vector "r" and "v" respectively:

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

v = (v0x, v0y + g*t)

Where:

r = position vector of the ball

x0 = initial horizontal position

t = time

y0 = initial vertical position

g = acceleration due to gravity

v = velocity vector

Considering the center of our system of reference as the point at which the ball left Serena´s racket, x0 and y0 = 0.

We know that at a time t = 1.21 s the y-component of the position vector must be 0 (see "r final" in the figure). Then:

y0 + v0y * t + 1/2 * g * t² = 0          y0= 0

v0y * 1.21 s + 1/2 * (-9.8 m/s²) * (1.21 s)² = 0

v0y = -(1/2 * (-9.8 m/s²) * (1.21 s)²) / 1.21 s

v0y = 1/2 * 9.8 m/s² * 1.21 s

v0y = 5.93 m/s

If we see in the figure the trajectory of the ball if there had been no gravity ("s"), we will notice that it is a stright line with a slope of:

Δy/Δx = (0.95m(y) + 2.63m(y)) / 11.7 m(x) = 0.31 m(y) / m(x)

This slope means that the ball will go up 0.31 m for every meter it goes right.

Then, if initially the ball goes up 5.93 m every second, it will go right

(5.93 m(y) * (1 m(x) / 0.31 m(y)) = 19.1 m(x). Then, v0x = 19.1 m/s

The vector initial velocity will be:

v0 = (19.1 m/s, 5.93 m/s)

magnitude of v0 =$|v0| = \sqrt{(19.1m/s)^{2}+(5.93m/s)^{2}}= 20.0 m/s$

Another way to solve this is by using the equation for velocity:

We know that when the ball passes over the net, the vertical velocity is 0. Then, we can calculate the time at which the ball passes over the net and use that time to obtain v0x from the equation for position, since we know that at that time the x-component of the position is 11.7 m.

When the ball is over the net:

v0y + g*t = 0

t = -v0y/g = -5.93 m/s/-9.8 m/s² = 0.61 s.

Notice that, since the trajectory is a parabola, knowing the final time we could easily calculate the time at which the ball passes the net by dividing that final time by 2: 1.21 s / 2 = 0.61 s

Then, using this time in the equation for position:

v0x * t = 11.7 m

v0x = 11.7 m / 0.61 s = 19.2 m/s which is aproximately the same as the obtained above.