Question

From the top of tower 100m in height a ball is dropped and at the same instant a ball is projected vertically upward with a velocity of 25m/s Find when and where the two balls will meet.(g=9.8m/s)

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

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

u = 0

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

Putting values,

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

$b2\quad \Rightarrow \quad d\quad =\quad x$

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

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

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

Putting values,

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

Adding (1) and (2),

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

t = 4 s

Putting t = 4 in (1),

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

$100\quad -\quad x\quad =\quad 78.4$

x = 80.4 m"

Answer 2

Answer:

Explanation:

100 - x = 78.4

∴ x = 21.6 m

The other answer is wrong