A rider on an open platform, which is descending at constant speed of 3 ms `s^(-1)`, throws a ball. Relative to platform, ball's initial velocity is horizontal at 12 m/s. The ground is 10m below the location where the ball is thrown:
(i) Where does the ball kit the ground?
(ii) How long after the ball hits the ground does the platform reach grond level?
(iii) With what velocity does the ball hit the ground?

Explanation

Due to inertia ball will share the velocity of plations at the instant of projection. Hence horizontal components of ball's velocity=12m/s and vertical components of ball's velocity =3m/s.
Considering the vertcial motion,
`10=3t+(1)/(2)xx9.8t^(@)`
`49t^(2)+30t-100=0`
`t=(-30+-sqrt(900+19600))/(98)`
Rejecting -ve root, we have
`t=(-30+10sqrt(205))/(98)=1.15s`(approx.)
(i) The distance AB=`12xx1.15=13.80m`
(ii) The time taken by platform to reach the ground =10/3=3.33 sec. Thus, the time difference=3.33-1.15=2.18s
(iii) Let the ball strike the ground with velocity v, then horizontal component `v_(h)=12m//s` and vertical component
`v_(v)^(2)=3^(2)+2xx9.8xx10`
`rArr v_(v)=14.32m//s`
Then `v=sqrt(v_(h)^(2)+v_(v)^(2))=sqrt(144+205)=sqrt(349)m//s`
`=18.68ms^(-1)`
and `tan theta=v_(v)//v_(h)=(14.32)/(12)=1.19`
`theta=tan^(-1)1.19`