Question

A ball is dropped from the top of a building 100 m high. at the same instant another ball is thrown upwards with a velocity of 40 m/s from the bottom of the building. the two balls will meet after

Answer 1

h= 100m
For ball 1: u=0; g=10m/s
So, h1=u1+1/2gt² =0×t+1/2(10)²t²= 5t²----------1
For ball2: u=40m/s;a= -10m/s
So, h2=u2+1/2at²=40×t+1/2(-10)²t²= 40t-5t²---------2
h=h1+h2
h=5t²+40t-5t² h=100m
100=40t
t=5/2=2.5 is answer

I HOPE THIS WILL HELP U

Answer 2

Answer:

The two balls will meet after 2.5 sec.

Explanation:

Given that,

Total height  = 100 m= s₁+s₂

Using equation of motion

For first ball

$s_{1} = ut+\dfrac{1}{2}gt^2$

Where, s = distance

u = initial velocity

g = acceleration due to gravity

t = time

$s_{1}=0-\dfrac{1}{2}gt^2$

For second ball

$s_{2} = 40t+\dfrac{1}{2}gt^2$

$s_{2}=40t-s_{1}$....(I)

Put the value of s₁ in equation (I)

$s_{2}+s_{1}=40t$

$100=40t$

$t = 2.5\ sec$

Hence, The two balls will meet after 2.5 sec.