Question

From the top of a tower $100m$ in flight ball is dropped and the same time another ball is projected vertically upward from the ground with a velocity of $25m/sec$. Find when and where the two balls will meet? $(g=9.8m/se{c}^2)$

Answer 1

Let take distance covered by the stone thrown upwards to meet the other as $x$. Then the other stone covers a distance of $(100-x)m$.

Let the ball dropped down be $b_1$. So the ball thrown up is $b_2$.

For $b_1$

$d=100-x$

$g=9.8$

$u=0$

Now, from equation of motion

$s=ut+\frac{1}{2}a{t}^2$

$100-x=\frac{1}{2}\times 9.8t^2$

$100-x=4.9t^2.....\left(I\right)$

For $b_2$

$g=-9.8$

$u=25m/s$

Now, from equation of motion

$s=ut+\frac{1}{2}g{t}^2$

$x=25t-4.9t^2....\left(II\right)$

Now, adding equation (I) and (II)

$100-x+x=4.9t^2+25t-4.9t^2$

$100=25t$

$t=4s$

Now, putting the value of t in equation (I)

$100-x=4.9\times 16$

$-x=78.4-100$

$x=21.6m$

Hence, they meet at $21.6m$ from ground after $4s$