# A tennis player hits a ball vertically with a speed of 10 m/s. How high does it go?

Nov 25, 2016

The ball reaches a height of 5.1 m.

#### Explanation:

The formula we are going to use is: $\left({V}_{\text{f"^2) = (V_"i}}^{2}\right) + 2 a d$.

=> where ${V}_{\text{f}}$ is the final velocity $\left(\frac{m}{s}\right)$, ${V}_{\text{i}}$ is the initial velocity $\left(\frac{m}{s}\right)$.
$a$ is the acceleration $\left(\frac{m}{s} ^ 2\right)$, and $d$ is the displacement $\left(m\right)$.

Acceleration is $9.8 \left(\frac{m}{s} ^ 2\right) \left[\mathrm{do} w n\right]$ , because that is the gravitational acceleration. It is negative because it is directional. I associate positive with $\left[u p\right]$, so negative is $\left[\mathrm{do} w n\right]$. It doesn't really matter which is which, as long as you're consistent.

$\left({V}_{\text{f"^2) = (V_"i}}^{2}\right) + 2 a \Delta d$

$0 = {10}^{2} + 2 \left(- 9.8\right) \Delta d$

$0 = 100 - 19.6 \Delta d$

$- 100 = - 19.6 \Delta d$

$5.1 = \Delta d$

The change in displacement is 5.1 m. Assuming the player is on the ground (0 m), that means the ball reaches a height of 5.1 m.

Hope this helps :)

Nov 25, 2016

The maximum altitude of the ball is $5.1 m .$

#### Explanation:

There are at least two ways to solve this, one being with kinematics and the other with energy conservation. These are just the first two that come to mind. I'll give an explanation of both methods.

Kinematics:

This problem exemplifies projectile motion. What is important to recognize here is that at its maximum altitude, the velocity of the ball will be $0$, as it will momentarily stop before falling back toward the earth. Also, we know that the acceleration ($y$-direction) is $- g$, or $- 9.8 \frac{m}{s} ^ 2$. We can use these known values for initial and final velocity and acceleration to determine the ball's maximum altitude:

(v_f)^2=(v_i)^2+2a_yΔy

Solving for Δy,

Δy=((v_f)^2-(v_i)^2)/(2a_y)

We know that ${v}_{f} = 0$

Δy=(-(v_i)^2)/(2a_y)

Δy=(-(10m/s)^2)/(2(-9.8m/s^2))

Δ=(-100m^2/s^2)/(-19.6m/s^2)

Δy=5.1m

Energy Conservation:

${U}_{g i} + {K}_{i} = {U}_{g f} + {K}_{f}$

Where ${U}_{g}$ is the initial gravitational potential energy (initial and final), and $K$ is the kinetic energy (initial and final).

As the ball is moving initially, for our intents and purposes it possesses only kinetic energy, the energy associated with motion (we can assign and initial height of $0$). When it reaches it's maximum height, it stops momentarily before falling back to the earth. At that point, the ball is not in motion, and possesses only gravitational potential energy. Therefore, our equation becomes:

${K}_{i} = {U}_{g f}$

This tells us that all of the energy the ball possesses initially in the form of kinetic energy is stored as potential energy when the ball is at its maximum altitude, which will then be converted back into kinetic energy when the ball is falling back to Earth.

Kinetic energy is given by $K = \frac{1}{2} m {v}^{2}$ and gravitational potential energy is given by ${U}_{g} = m g h$. Thus, we have:

$\frac{1}{2} m {v}^{2} = m g h$

We can see that mass cancels, yielding:

$\frac{1}{2} {v}^{2} = g h$

Solving for $h$,

$h = \frac{\frac{1}{2} {v}^{2}}{g}$

Using ${v}_{i} = 10 \frac{m}{s}$, ${v}_{f} = 0$, and $g = 9.8 \frac{m}{s} ^ 2$

$h = \frac{\frac{1}{2} {\left(10 \frac{m}{s}\right)}^{2}}{9.8 \frac{m}{s} ^ 2}$

$h = \frac{50 {m}^{2} / {s}^{2}}{9.8 \frac{m}{s} ^ 2}$

$h = 5.1 m$

Hope that helps!