Question

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
velocity of 25 ms. Find when and where the two
balls will meet. Take g = 9.8 ms [Delhi 05]
(Ans. 78.4 from top, 4 s)​

Answer 1

Answer:They will meet instinctively. Here's how.

Lets 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 ball dropped down be b1. So the ball thrown up is b2.

b1 => d = 100-x

g = 9.8 m/s sq.

u = 0

We know, s= ut + 1/2at sq.

Putting values,

100-x = 4.9t sq. ....(1)

b2 => d = x

g = -9.8 m/s sq.

u = 25 m/s

We know, s= ut + 1/2at sq.

Putting values,

x = 25t -4.9t sq. ....(2)

Adding (1) and (2),

100-x +x = 4.9t sq. + 25t - 4.9t sq.

t = 4 secs.

Puuting t = 4 in (1),

100-x = 4.9t sq.

100-x =19.6

x = 80.4 m

So, they meet at 80.4 m from ground after 4 seconds.

Please mark as brainliest

Answer 2

Explanation:

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

velocity of 25 ms. Find when and where the two

balls will meet. Take g = 9.8 ms [Delhi 05]

(Ans. 78.4 from top, 4 s).