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)`)

Options

A. `2.396 m`
B. `3.396 m`
C. `4.396 m`
D. `5.396 m`

Answer

B

Explanation

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^(@)) xx 3 - (10)/(2(100 cos^(2) 60^(@))) (3)^(2)`
`=3 sqrt(3) - (9)/(5) = (15 sqrt(3)-9)/(5) m = 3.396 m`