Question

A stone is allowed to fall from the top of a tower $100\hspace{0.17em}\text{m}$ high and at the same time, another stone is projected vertically upwards from the ground with a velocity of $25\frac{\text{m}}{\text{s}}$. Calculate when and where the two stones will meet.

Answer 1

Step 1: Given data

Initial velocity for upper stone, $u\text{'}=0$.

Initial velocity for lower stone, $u=25\frac{\text{m}}{\text{s}}$.

Height, $h=100\text{m}$.

Step 2: To find

When and where the two stones will meet.

Step 3: Calculate time and position of meeting

Let the stones meet at point A after time $t$.

Displacement for upper stone:

$$\begin{gathered} x=u\text{'}t+\frac{1}{2}g{t}^2 \\ x=\left(0\right)t+\frac{1}{2}\times 10\times t^2 \\ x=5t^2 \end{gathered}$$

Displacement for lower stone:

$$\begin{gathered} 100-x=ut-\frac{1}{2}g{t}^2 \\ 100-x=\left(25\right)t-5t^2 \end{gathered}$$

Adding displacement for upper stone and displacement for lower stone:

$$\begin{gathered} 100=25t \\ t=4\text{s} \end{gathered}$$

Substitute $4\text{s}$ for $t$ in displacement for upper stone:

$$\begin{gathered} x=5t^2 \\ =5{\left(4\right)}^2 \\ =80\text{m} \end{gathered}$$

Hence, the stone meet at a height of $100\text{m}-80\text{m}=20\text{m}$ above the ground after $4\text{s}$.