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

To calculate the speed of the ball when it left Serena's racket, we can use the principle of conservation of energy. The initial potential energy of the ball is equal to its final kinetic energy. Calculating the above values, we find that the speed of the ball when it left Serena's racket was approximately 46.7 km/h.

Step-by-step explanation:

To calculate the speed of the ball when it left Serena's racket, we can use the principle of conservation of energy. The initial potential energy of the ball is equal to its final kinetic energy. In this case, the initial potential energy is the energy required to raise the ball 2.63 m higher than the net, and the final kinetic energy is the energy the ball has when it leaves Serena's racket.

We can start by calculating the initial potential energy using the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height. Given that the height is 2.63 m and the mass of a standard tennis ball is around 0.057 kg, we find that the initial potential energy is PE = 0.057 kg * 9.8 m/s^2 * 2.63 m.

Next, we can calculate the final kinetic energy using the formula KE = 0.5 * mv^2, where m is the mass of the ball and v is its velocity. Rearranging the formula, we can solve for v: v = sqrt(2 * KE / m). Given that the final kinetic energy is equal to the initial potential energy, we can substitute the values and solve for v. Finally, we can convert the velocity from m/s to km/h by multiplying by 3.6.

Calculating the above values, we find that the speed of the ball when it left Serena's racket was approximately 46.7 km/h.

Answer 2

The initial velocity of the ball was 20 m/s

Step-by-step 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.