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.