This problem aims to find the maximum height of a golf ball that has been hit in a** projectile** manner at an angle of $25.0$ and covering a range of $305.1 m$. This problem requires the knowledge of **projectile displacement formulas,** which include **projectile** **range** and **height**.

**Projectile motion** is the term for the movement of an **object flung** or cast into the air, related to only the **acceleration** due to **gravity.** The object that is flung is known as a **projectile,** and its route is known as its course. This problem can be cracked using the equations of **projectile motion** with constant acceleration. As the object is covering a horizontal distance, the acceleration here must be null. Thus, we can express the **horizontal displacement** as:

\[ x = v_x \times t \]

Where $v_x$ is the horizontal component of the velocity and $t$ is the **flight time**.

## Expert Answer

We are given the following parameters:

$R = 301.5 m$, $R$ is the **horizontal distance** that the ball travels after a projectile motion.

$\theta = 25$, $\theta$ is the **angle** with which the ball is displaced from the ground.

The formula of vertical motion can be derived from the **first equation of motion**, which is given as:

$v = u + at$

where,

$v$ is the **final velocity**, and its value is the vertical component of the initial velocity –> $usin\theta$

$u$ is the **Initial velocity =** $0$

$a$ is the **negative Acceleration**, as the ball is moving **upwards** against the **force** of **gravity** = $-g$

The formula for **acceleration** due to gravity is $g = \dfrac{v – u}{t}$

Rearranging the above formula for value of $t$,

\[t=\dfrac{usin\theta}{g} \]

The formula for the **horizontal range** of **Projectile** motion is given:

\[R=v \times t \]

Plugging in the expressions of $v$ and $t$ gives us:

\[R=usin\theta \times \dfrac{usin\theta}{g} \]

\[ R=\dfrac{u^2 sin^2\theta}{g} \]

Now that we have our formula to compute the **final velocity,** we can further plug in the values to compute $u$:

\[301.5 = \dfrac{u^2 sin^2(25)}{9.8} \]

\[\dfrac{301.5 \times 9.8}{sin^2(25))} = u^2 \]

\[u^2 = 3935 m/s \]

Next, to compute the **maximum height **of the projectile $H$, we will use the formula as given:

\[H = \dfrac{u^2 sin^2\theta}{2g} \]

\[H = \dfrac{3935 \times sin^2(25)}{2(9.8)} \]

## Numerical Result

The **maximum height** is calculated to be:

\[H = 35.1 m \]

## Example:

A **golfer hits** one **golf ball** at an **angle** of $30^{\circ}$ to the ground. If the golf ball covers a **horizontal distance** of $400$, what is the ball’s **maximum altitude?**

The formula for the **horizontal range** of **Projectile motion** is given:

\[R = \dfrac{u^2 sin^2\theta}{g} \]

Now that we have our formula to compute the **final velocity,** we can further plug in the values to compute $u$:

\[400 = \dfrac{u^2 sin^2(30)}{9.8} \]

\[\dfrac{400 \times 9.8}{sin^2(30))} = u^2\]

\[u^2= 4526.4 m/s\]

Finally, to compute the **maximum height** of the **projectile** $H$, we will use the formula as given:

\[H=\dfrac{u^2 sin^2\theta}{2g}\]

\[H=\dfrac{4526.4 \times sin^2(30)}{2(9.8)}\]

**Horizontal distance** comes out to be:

\[H = 57.7 m\]

*Images/mathematical drawings are created with GeoGebra*