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A stone is thrown with a speed of 10 ms^...

A stone is thrown with a speed of `10 ms^(-1)` at an angle of projection `60^(@)`. Find its height above the point of projection when it is at a horizontal distance of 3m from the thrower ? (Take `g = 10 ms^(-2)`)

Text Solution

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Considering the equation of trajectory.
` y = ( tan theta_(0)) x - g/(2 (v_(0)^(2) cos^(2) theta_(0))) x^(2)`
Here ` theta_(0) = 60^(@)`
` v_(0) = 10 ms ^(-1)`
x =3m
` y= ( tan 60^(@))3 - 10/(2 (100 cos^(2) 60^(@)) (3)^(2)`
` = 3 sqrt3 - 9/5 `
` ( 15 sqrt3 -9)/ 5 ` m
= 3.396 m
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