Physics, asked by bestiam938, 2 months ago

A ball is dropped from the top of a tower 100 m high and at the same time another
ball is projected vertically upwards from the ground with a velocity of 25 m/sec. Calculate
where and when the two stones will meet. (Take g = 9.8 ms)

Answers

Answered by piyush000880
0

Answer:

For ball 1:

u

1

=0 a

1

=g=10m/s

2

s

1

=100−x

S

1

=u

1

t+

2

1

a

1

t

2

⇒ 100−x=

2

1

(10)t

2

....(1)

For ball 2:

u

1

=−25m/s a

2

=g=10m/s

2

s

2

=−x

S

2

=u

2

t+

2

1

a

2

t

2

⇒ −x=−25t+

2

1

10t

2

.

⇒ x=−25t−5t

2

....(2)

Step 3: Solving equations

From equation (1) and (2)

100−25t+5t

2

=5t

2

⇒ t=4s

Putting the value of t in equation (1)

x=100−5×(4)

2

=20m

Hence they will meet 20m above the ground after t=4s

Explanation:

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