A ball projected with a velocity of `10ms^(-1)` at an angle of `30^(@)` with the horizontal just clears two vertical poles, each of height 1.0 m. Find the separation between the poles. Take `g=10ms^(-2)`

Explanation

Refer to figure. Let the calculate the two values of t at which the ball passes just above P and R. For each pole h=1.0 m

y-component of velocit is `u_(u),u_(u)=usin theta=10sin30^(2)=5ms^(-1)`
`h=u_(y)=1/2"gt"^(2)`
`implies1.0=5t+1/2(-10)t^(2)`
`=5t^(2)-5t+1.0=0`
The two roots of this quandratic equation are `t_(1)=0.72` and `t_(2)=2.76s`. Therefore
`OQ=u_(x)t_(1)=10cos 30^(@)xx0.72=6.2m`
and `OS=u_(x)t_(2)=10cos 30^(@)xx2.76=23.9m" ":.QS=23.9-6.2=17.7m`