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Grade: 10

                        

The speed of a projectile when it is at its greatest height is (2/5)raised to the power 1/2 times its speed at half the maximum height.The angle of projection is? Detailed answer I want

5 years ago

Answers : (1)

Vikas TU
12279 Points
							
given,
v = (2/5)^(0.5)*v’
at max. height the speed is,
H = v’^2/2g
or
v’ = root(2gH)
at half . of max. height  =>
H/2 = 2gHsin^2thetha/2g
solve for thetha.
4 years ago
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