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a particle is projected with speed u at an angle of elevation alpha, where tan alpha = 3. find in terms of u and g the height of the particle
Let the max height attained be h. At highest point of trajectory vy=0 From third eqn of motion, analysing motion in vertical dirxn onlyv2=u2-2ghu2/2g=h=(u sin alpha)2/2g Now since tan alpha = 1/3sin alpha=1/{(12+32)1/2} = 1/(10)1/2 h=u2/(10x2g) = u2/20g
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