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A jet of water is projected at an angle ...

A jet of water is projected at an angle `theta = 45^@` with horizontal from point `A` which is situated at a distance `x = OA = (a) 1//2 m, (b) 2 m` from a vertical wall. If the speed of projection is `v_0 = sqrt(10 ms^-1`, find point `P` of strinking of the water jet with the vertical wall.
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Text Solution

Verified by Experts

Point `P` lies at the trajectory of jet of water. Hence, the coordinate of point `P (x,y)` should satisfy the trajectory equation.
`y = x tan theta (1 - (x)/(R ))`…(i)
`R = (u^2 sin 2 theta)/(g) = ((sqrt(10))^2 . sin 2 xx 45^@)/(10) = 1 m`
(a) If `x = 1//2 m` from (i) `y = (1)/(2) tan 45^@ (1 - (1//2)/(1)) = (1)/(2) xx (1)/(2) = (1)/(4) m`
Hence, the coordinate of `P = ((1)/(2) m, (1)/(4) m))`
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(b) If `x = 2m`, from (i),
`y = 2 . tan 45^@ (1 - (2)/(1)) = -1 m`
Hence,water jet will strike below the horizontal dotted line (x - axis) at coordinate `(2m, -1m)`.
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