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A helicopter while flying at a height of...

A helicopter while flying at a height of 100 m with velocity 30 m.s at an angle `30^(@)` with the horizontal, drops a packet. Where will the packet strike the ground ? ( g = 10 ` m//s^(2)`)

Text Solution

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Initial velocity of the packet will be same as that of the helicopter. Therefore for the packet intialiy.

`u_(x) = u cos 30^(@) = 30 sqrt3/2 = 15 sqrt3` m/s
` u_(y) = u sin 30^(@) = 15( 1/2) = 7.5 `
`a_(x) =0`
` a_(y) = -10 m//s^(2)`
Vertical motion
When it strikes the ground its displacement in vertical direction is (-h) = ( -100 m).Now, form equation.
` S = ut + 1/2 at^(2) , ` we get
` -h = u sin theta t - 1/2 gt^(2)`
` - 100 = 15t - 1/2 (10)t^(2)`
` Rightarrow 5t^(2) - 15t - 100 =0`
` Rightarrow t^(2) - 3t - 20 =0`
On solving this equation for positive root to t
` t = ( 3 + sqrt( g + 80))/2 = 6.23 ` second.
It packet is released above the point P on the ground and it strikes the ground at point Q then the distance between P and Q.
s = u_(x) t
` = 15sqrt2 xx 6.23 m`
132 . 14 m
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