Question

A ball is dropped from the top of a tower 100 metre in height and at the same time another ball is projected vertically upwards so as to just reach at the top of the tower. Find when and where the two balls will meet. g = 9.8. Velocity is not given.

Answer 1

Answer:

Am I

Explanation:

i guess Form the top of a tower 100m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velcoity of 25ms-1. ... Hence, the two balls will meet after 4seconds at a distance 78.4m below the top

Answer 2

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

.........(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

It is a example of the question

change the value and u will get the answer

I hope so