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 us calculate the two values of t at which the ball passes just above P and R. for each pole h=1.0 m
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y-component of velocity is `u_(y),u_(y)=usintheta=10sin30^(@)=5ms^(-1)`
`h=u_(y)t+(1)/(2)g t^(2)` `implies 1.0=5t+(1)/(2)(-10)t^(2)`
`implies 5t^(2)-5t+1.0=0`
The two roots of this quadratice equation are `t_(1)=0.72 s and t_(2)=2.76s`. therefore
`OQ=u_(x)t_(1)=10cos30^(@)xx0.72=6.2m`
and `OS=u_(x)t_(2)=10cos30^(@)xx2.76=23.9m" "thereforeQS=23.9-6.2=17.7m`