Question

A ball is dropped from a height of 40 m. At the
same time another ball is thrown upward with a
velocity of 20 m/s. Both the balls with meet after please help​

Answer 1

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^2s

1

=ut+

2

1

gt

2

Where, s = distance

u = initial velocity

g = acceleration due to gravity

t = time

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

1

=0−

2

1

gt

2

For second ball

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

2

=40t+

2

1

gt

2

s_{2}=40t-s_{1}s

2

=40t−s

1

....(I)

Put the value of s₁ in equation (I)

s_{2}+s_{1}=40ts

2

+s

1

=40t

100=40t100=40t

t = 2.5\ sect=2.5 sec

Hence, The two balls will meet after 2.5 sec.

hope this is helpful to you