Question:

A ball is dropped from height 100 m and another ball is projected vertically up with a velocity 50 ms^-1 from the ground along the same line. Find out the position where the two balls will meet?

Solution:

First case, when the first ball is dropped

Motion under gravity- Sign convention

$${100-y=\frac{1}{2}gt^2}$$

$${y=100-\frac{1}{2}gt^2}$$

Let it be h₁(t)

Second case: ball 2 is thrown up

$${y=ut-\frac{1}{2}gt^2}$$

$${y=ut-\frac{1}{2}(10)t^2}$$

$${y=ut-\frac{1}{2}gt^2}$$

$${y=50t-\frac{1}{2}gt^2}$$

Let it be h₂(t)

To find the position where the two balls meet h₁(t) =h₂(t)

$${100-\frac{1}{2}gt^2=50t-\frac{1}{2}gt^2}$$

$${50t=100}$$

$$t=2 \ sec$$

Put t=2 s in h₂(t)

$${h_2(t)=100-\frac {1}{2}(10)(2^2)}$$

$${h_2(t) = 80 \ m}$$

The two balls will meet at apoint 80 m from the ground.