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A projectile is thrown with speed ‘v’ at an angle ‘x’ with upward vertical.It’s average velocity between the instants at which it crosses half of maximum height is?
Maximum height is given as:H = u^2/2gand half of its maximum height is: H/2 = > u^2/4gNow u need to apply conservation of energy to find te veloocity at H/2.U avg. = ucosthetha. i thinnk so.
H/2= V2 Sin2theta/4g:- now you can find by placing values....So, according to me answer is..Vsintheta , horizontal and in the plane of projections
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