From the top to a tower ` 100 m` in height a ball is dropped and at the same instant another ball is projeced veticall y upwards from the ground so that it just reaches the top of tower. At what height do the two balls pass one another ?

Explanation

Taking vertical upward motion of ball from ground to heith ` 100m`, we have
` u=? A=-9.8 m//s^2, v=0 , S= 100 m` ltbRgt As, ` v^2 =u^2 + 2 aS,
so `0=u^2 + 2(- 9.8) xx 1000
or ` u = sqrt 9 2 xx 9. xx 100) = 14 sqrt 10 m//s`
Let the two balls meet ater time (t) and at a heitht (x) metres from ground. The vertical distance travelled by ball dropped from ` 100 m` in time (t) ` = ( 100 - x0 metres.
Taking vertical downward motion of one ball for time (t0, we have
` u=0, a = 9.8 m//s^2 S= ( 1200 - x) , t=t`
As, `Sput 1/2 at^2`
:. 9 100-x) =0 0= 1/2 xx 9.8 t^2 = 4. 9 t^2 ` ..(i)`
Taking vertcal upward motion of anlther ball for time (t), we have
` u 14= sqrt 10 m//s , a=- 9.8 m//s^2 , t=t , S= x`
As, ` S=ut + 1/2 at^2 , So`
` x = 14 sqrt 10 t + 1/2 ( -9.8 0 t^2`
` = 14 sqrt 10 t - 4.9 t^2`
Adding (i) and (ii) we get ...(ii)
` 100 =14 sqrt 10 t or t= ( 100) /( 14 sqrt 10 ) = ( 10 sqrt 10)/ ( 14)`
For, (ii) `
` x = 14 sqrt 10 xx ( 10 sqrt 10) / ( 14) - 4.9. ( ( 10)m /( 14) xx sqrt10 )^@`
` = 100 - 25 - 75 m`