**How long after the release of the first stone does the second stone hit the water?****What was the initial speed of the second stone?****What is the speed of each stone as it hits the water?**

This question aims to find the **time** of the **stone** as it **hits** the **water,** the **initial speed** of the **second stone,** and the **final speed** of **both** **stones** as they hit the water.

The basic concepts needed to understand and solve this problem are **equations of motion, gravitational acceleration,** and **initial** and **final velocities** of an object during **vertical fall.**

## Expert Answer

We are taking the **initial point** at the **cliff** as the starting point, hence the **final height** will be at the **water surface** and the **initial height** will be at the **cliff.** Also, the **downward motion** will be taken as **positive.**

The given information regarding this problem is given as follows:

\[ The\ Initial\ Velocity\ of\ the\ First\ Stone\ v_i\ =\ 2.5\ m/s \]

\[ The\ Final\ Height\ h_f\ =\ 70\ m \]

\[ The\ Initial\ Height\ h_i\ =\ 0\ m \]

\[ The\ Acceleration\ due\ to\ Gravity\ g\ =\ 9.8\ m/s^2 \]

**a)** To calculate the **time** the **second stone** took to hit the water after the **first stone,** we will use the equation of motion, which is given as:

\[ h_f = h_i + v_it + \dfrac{1}{2} at^2 \]

Substituting the values, we get:

\[ 70 = 0 + 2.5t + \dfrac{1}{2} (9.8) t^2 \]

\[ 4.9t^2 + 2.5t – 70 = 0 \]

By using the **quadratic formula,** we can calculate the value of $t$, which is calculated to be:

\[ t_1 = 3.53\ s \]

Ignoring the **negative value** of $t$ as time is always positive.

The **second stone** was released $1.2s$ after the **first stone** was released, but reached the water at the **same time.** So the time the **second stone** took to reach the water is given as:

\[ t_2 = 3.53\ -\ 1.2 \]

\[ t_2 = 2.33\ s \]

**b)** To calculate the **initial velocity** of the **second stone,** we can use the same equation. The initial velocity can be calculated as:

\[ h_f = h_i + v_it_2 + \dfrac{1}{2} gt_{2}^{2} \]

Substituting the values, we get:

\[ 70 = 0 + v_{i2} (2.33) + (0.5 \times 9.8 \times (2.33)^2 \]

\[ v_{i2} = \dfrac{70 – 26.6} {2.33} \]

\[ v_{i2} = \dfrac{43.4}{2.33} \]

\[ v_{i2} = 18.63\ m/s \]

**c)** To calculate the **final velocities** of **both stones,** we can use the following **equation** of **motion:**

\[ v_f = v_i + gt \]

The **final velocity** of the **first stone** is given as:

\[ v_{f1} = 2.5 + 9.8 \times 3.53 \]

\[ v_{f1} = 37.1\ m/s \]

The **final velocity** of the **second stone** is given as:

\[ v_{f2} = 18.63 + 9.8 \times 2.33 \]

\[ v_{f2} = 41.5\ m/s \]

## Numerical Results

**a)** The **total time the second stone** took to hit the water:

\[ t_2 = 2.33\ s \]

**b)** The **initial velocity of the second stone** is calculated as:

\[ v_{i2} = 18.63\ m/s \]

**c)** The f**inal velocities of both stones** are calculated as:

\[ v_{f1} = 37.1\ m/s \hspace{0.6in} v_{f2} = 41.5\ m/s \]

## Example

The **initial velocity** of an object is $2m/s$ and it took the object $5s$ to reach the **ground.** Find its **final velocity.**

As the object is **falling,** we can take the **acceleration** $a$ to be the **gravitational acceleration** $g$. By using the first **equation** of **motion,** we can calculate the **final velocity** without knowing the **total height.**

\[ v_f = v_i + gt \]

\[ v_f = 2 + 9.8 \times 5 \]

\[ v_f = 51\ m/s \]

The **final velocity** of the object is calculated to be $51 m/s$.