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Grade 11Mechanics

The total speed of a projectile at its greatest height is √(6/7) of its speed when it is at half of its greatest height . The angle of projection will be?

kruti

9 Years agoGrade 11

1 Answer

Vikas TU

9 Years ago

Let greatest height be ‘H’.

then,

H = u^2/2g

at half of its greatest height let

h = v^2sin^2theta/2g

Now, h = H/2 {given}..............(1)

and u = √(6/7)*v............(2)

Putting both eqns. in H and h and solving after we get,

2/1 = (6/7)sin^2(thetha)

sin(thetha) = root(3/7)

thetha = 40.84 degree.

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