Question

an object is fall from a tower 100 M with time 1 second find the velocity when object striking on the ground

Answer 1

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.

  

Answer 2

use v^2=u^2+2as

v^2=u^2+2(g)(100)
v^2=+u^2+200g.....eqn1.

v=u+g(1)
v=u+g......eqn 2

2EQNS 2 VARIABLES(v and u).
solve them to get the answer.

let me knw if u hv any doubt.