Question

please answer it quick it's urgent From the top of a tower 100 m in height a ball is dropped and at the same time
another ball is projected vertically upwards from the ground with a velocity of 25 m/s.
Find when and where the two balls will meet. (g = 9.8 m/s)
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Answer 1

The two balls will meet at 80.4 m from ground after 4 seconds.

Solution:

Let us take distance covered by the stone thrown upwards to meet the other as x. Then the other stone covers a distance of (100 - x) m

Let the distance covered by the ball dropped down be b1 and the distance covered by the ball thrown up be b2.

b1 = d = 100-x

g\quad =\quad 9.8\quad m/s^{ 2 }g=9.8m/s

2

u = 0

We know, s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } at^{ 2 }s=ut+

2

1

at

2

Putting values,

100\quad -\quad x\quad =\quad 4.9t^{ 2 }\quad \longrightarrow (1)100−x=4.9t

2

⟶(1)

b2\quad \Rightarrow \quad d\quad =\quad xb2⇒d=x

g\quad =\quad 9.8\quad \frac { m }{ s^{ 2 } }g=9.8

s

2

m

u\quad =\quad 25\quad \frac { m }{ s }u=25

s

m

We know, s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } at^{ 2 }s=ut+

2

1

at

2

Putting values,

x\quad =\quad 25t\quad -\quad 4.9t^{ 2 }\quad \longrightarrow (2)x=25t−4.9t

2

⟶(2)

Adding (1) and (2),

100\quad -\quad x\quad +\quad x\quad =\quad 4.9t^{ 2 }\quad +\quad 25t\quad -\quad 4.9t^{ 2 }100−x+x=4.9t

2

+25t−4.9t

2

t = 4 s

Putting t = 4 in (1),

100\quad -\quad x\quad =\quad 4.9t^{ 2 }100−x=4.9t

2

100\quad -\quad x\quad =\quad 78.4100−x=78.4

x = 80.4 m"