Question

A ball is the thrown vertically upward with an initial velocity 100 m per sec., simultaneously 2nd ball dropped from a height 200m ,find at what height from the ground and what time the two ball meet​

Answer 1

They will meet at distance of 80 m from the ground after t = 4 seconds

Answer 2

Answer:

$.$

Explanation:

Let the ball dropped down is a and the ball thrown up is b. also the stone thrown upwards cover a distance of x and the other covers a distance of (100 –x).

For ball a

u=0

g=10m/s

2

d=(100−x)

Using the equation

s=ut+

2

1

at

2

100−x=5t

2

.

For ball b

d=x

g=−10m/s

2

u=25m/s

s=ut+

2

1

at

2

x=25t−5t

2

.

Solving equation (1) and (2)

100=25t

t=4seconds

Put the value of t in equation (1)

x=100−80

x=20m

They will meet at distance of 80 m from the ground after t = 4 seconds