Question

From the top of a tower $100m$ in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with velocity of $25m{s}^{-1}$. Find when and where the two balls will meet. Take $g=10m{s}^{-2}$

Answer 1

$\text{Step 1: Analysing the problem [Refer Figure]}$

Let the two ball meet at point $C$ at time t

If the distance covered by ball 2 is $x$, then the distance covered by ball 1 will be $100-x$, Since the total hieght is $100m$.

$\text{Step 2: Equations of motion for constant acceleration}$

Since acceleration is constant, Therefore we can apply equations of motion for both balls (considering Positive Downwards).

$\text{For ball 1:}$

$u_1=0$ $a_1=g=10m/s^2$ $s_1=100-x$

$S_1=u_{1}t+\frac{1}{2}{a}_1{t}^2$

$\Rightarrow$ $100-x=\frac{1}{2}\left(10\right)t^2$ $....\left(1\right)$

$\text{For ball 2:}$

$u_1=-25m/s$ $a_2=g=10m/s^2$ $s_2=-x$

$S_2=u_{2}t+\frac{1}{2}{a}_2{t}^2$

$\Rightarrow$ $-x=-25t+\frac{1}{2}10t^2$ .

$\Rightarrow$ $x=-25t-5t^2$ $....\left(2\right)$

$\text{Step 3: Solving equations}$

From equation $\left(1\right)$ and $\left(2\right)$

$100-25t+5t^2=5t^2$

$\Rightarrow$ $t=4s$

Putting the value of $t$ in equation $\left(1\right)$

$x=100-5\times (4{)}^2$

$=20m$

Hence they will meet $20m$ above the ground after $t=4s$