Question

A ball is thrown vertically upward with 50m/s and at the same time another ball is dropped from the height of 100 m . Find the time when they meet with each other.​

Answer 1

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Explanation:

Answer 2

$\bigstar$ Solution :

$\bigstar$ Question :

A ball is thrown vertically upward with 50 m/s and at the same time another ball is dropped from the height of 100 m . Find the time when they meet with each other.​

$\bigstar$ Explanation :

So ,

$u_1=50m/s,s_1=100-x,a_1=-g,u_2=0,s_2=x$

For First Ball ,

$s_1=u_1t_1+\frac{1}{2}a_1t_1^2\\\\ \implies (100-x)=50.t+\frac{1}{2}.(-g).t^2 \\\\\implies (100-x)=50t+\frac{-gt^2}{2}...(1)$

For Second Ball ,

$s_2=u_2.t_2+\frac{1}{2}.a_2.t_2^2\\\\ \implies x=(0)t+\frac{gt^2}{2}\\\\ \implies x=\frac{gt^2}{2}...(2)$

Now , add (1) and (2) , we get

$\implies (100-x)+x=50t-\frac{gt^2}{2}+\frac{gt^2}{2}\\\\\implies 100=50t\\\\\implies t=2 seconds$

                                                         

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