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

Question

Agrata Singh · Answer

Let the max height attained be h. At highest point of trajectory v y =0 From third eqn of motion, analysing motion in vertical dirxn only v 2 =u 2 -2gh u 2 /2g=h=(u sin alpha) 2 /2g Now since tan alpha = 1/3 sin alpha=1/{(1 2 +3 2 ) 1/2 } = 1/(10) 1/2 h=u 2 /(10x2g) = u 2 /20g

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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

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

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