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# The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x .The angle of projection with the horizontal is ?

Asked by Ayradhey 30th October 2018, 9:43 PM
Let u be the velocity of projection and α is the angle of projection (angle made by velocity vector with horizontal).

At top of the trajectory, projectile has only horizontal component of velocity u×cosα .

Hence ratio of the projection velocity to the velocity at top most point = u / ( u×cosα )  = secα ............(1)

This ratio is given as x .  Hence the required ratio = secα = x .............(2)

from eqn.(2), we can write angle of projection α = cos-1 (1/x)
Answered by Expert 31st October 2018, 1:02 AM
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