Question

a ball is allowed to fall from the top of a tower 200 m high and at the same time another ball is projected vertically upwards from the ground with velocity of 25m/s. Calculate when and where the two balls will meet.​

Answer 1

Answer:

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

.........(1)

For ball b

d=x

g=−10m/s

2

u=25m/s

s=ut+

2

1

at

2

x=25t−5t

2

............(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

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