This problem aims to find the **acceleration due** to the **gravity** of an object on a **distant planet.** The concepts required to solve this problem are related to **gravitational physics,** which include **Newton’s equations of gravitational motion.**

A **motion** under the influence of **gravity** directs to the **vertical** movement of an object whose motion is impacted by the existence of **gravity.** Whenever an object falls, a **force** attracts that object **downwards** known as **gravity.**

**Newton’s equations** of motion are related to an object moving in a **horizontal direction,** which means there is no **gravitational acceleration** imposed on the object, but if the object covers a **vertical distance, gravity** will occur and its equations are given as follows:

\[ v_f = v_i + at….\text{horizontal motion}\implies \space v_f = v_i + gt….\text{vertical motion} \]

\[ S = v_it + \dfrac{1}{2}at^2….\text{horizontal motion}\implies \space H = v_it + \dfrac{1}{2}gt^2….\text{vertical motion} \]

\[ 2aS = v^{2}_{f} – v^{2}_{i}….\text{horizontal motion}\implies \space 2gS = v^{2}_{f} – v^{2}_{i}….\text{vertical motion} \]

Where $H$ is the **height** of the **object** from the ground, $g$ is the **gravitational acceleration **acting on the **object,** and its value is $9.8 m/s^2$.

## Expert Answer

We are given the following **information:**

- The
**initial velocity**is with which the**rock**is thrown $v_i = 15\space m/s$, - The
**time**it takes for the rock to**reach back**$t = 20\space s$, - The
**initial location**of the rock $x = 0$.

Now we are going to take help from the** second equation of motion** under **gravity:**

\[ x = v_it + \dfrac{1}{2}gt^2\]

**Plugging** in the values:

\[ 0 = 15\times 20 + \dfrac{1}{2}(a)(20)^2\]

\[ 15\times 20 = -\dfrac{1}{2}(400a)\]

\[ 300 = -200a \]

\[ a = -\dfrac{300}{200} \]

\[ a = -1.5\space m/s^2 \]

Therefore, the **acceleration** is of **magnitude** $1.5\space m/s^2$ and the **negative** sign indicates that the **direction** of motion is **downward.**

## Numerical Result

The **acceleration** comes out to be of **magnitude** $1.5\space m/s^2$ and the **negative** sign here indicates that the **direction** of **motion** is **downward.**

## Example

The **player** kicks the **football** $25.0m$ from the **goal,** with the **crossbar** $8.0m$ high. The **speed** of the ball is $20.0 m/s$ when it leaves the **ground** at an **angle** of $48^{\circ}$ **horizontally,** how long does the ball **stay** in the **air** before reaching the **goal** area? How **far** does the ball **land** from the **crossbar?** And does the **ball reach** the crossbar while **going up** or falling **down?**

Since the ball is **moving** in the **horizontal** direction, the **velocity component** would look like this:

\[v_{0x} = v_0\cos \theta \]

And the **distance formula:**

\[\bigtriangleup x = v_{0x} t\]

**Rearranging:**

\[t= \dfrac{\bigtriangleup x}{v_{0x}}\]

\[t= \dfrac{25.0 m}{20.0 \cos (48)}\]

\[t= 1.87\space s\]

To find the **vertical distance** of the ball:

\[y=v_0\sin\theta t – \dfrac{1}{2}gt^2\]

\[y=20\sin (48) (1.87) – \dfrac{1}{2}(9.8)(1.87)^2\]

\[y=10.7\space m\]

Since the ball is $10.7m$ high, it **clears** the **crossbar** by:

\[10.7m-8.0m=2.7m\space\text{clears!}\]

To find the **rise** or **fall** of the ball while it approaches the **crossbar:**

\[v_y=v_0y – gt\]

\[v_y=v_0\sin\theta – gt\]

\[v_y=20\sin(48) – (9.8)1.87\]

\[v_y=-3.46\space m/s\]

The **negative sign** tells that it is **falling.**