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Grade: 11

                        

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?

4 years ago

Answers : (1)

Vikas TU
12273 Points
							
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.
4 years ago
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