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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)`) |
Answer» 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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