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Grade:
        
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
one year ago

Answers : (1)

Agrata Singh
208 Points
							
Let the max height attained be h. At highest point of trajectory vy=0 
From third eqn of motion, analysing motion in vertical dirxn only
v2=u2-2gh
u2/2g=h=(u sin alpha)2/2g
 
 
Now since tan alpha = 1/3
sin alpha=1/{(12+32)1/2} = 1/(10)1/2
 
h=u2/(10x2g) = u2/20g
one year ago
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