Question

The height of the tower is $200m$ a ball thrown up with a velocity of $50m/s$ , another ball dropped from a top of a tower, when and where they meet?

Answer 1

The distance travelled by the body dropped from the height is given as

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

$s=50t-5t^2$

Both the body will together travel the distance of $200m$

Hence

$200=\left(50t-5t^2\right)+5t^2$

Time taken is$4\sec$

Distance travelled by the falling stone is

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

$s=5\times 8=80m$

The distance travelled by the projected body is

$s=50\times 4-5\times 4^2$

$s=120m$

So both body will meet at $120m$ above the ground.