an object is projected from ground with speed 20m/s at an angle 30 with horizontal .its centripetal acceleration 1s after the projection is

Dear Student,

Please find below the solution to the asked query:

Given that the initial velocity of the particle is 20 m/s and the angle of projection is 300. Let the velocity of the object after t = 1 sec is v. Therefore,

v cos α=u cos θv cos α=20 cos 30v cos α=103v sin α=v2-300&v sin α=u sin θ-gt20 sin 30-10×1=v2-300v2-300=0v=103 ms

So, the particle will be at its highest point. Therefore the centripetal acceleration is g as shown in the figure below.

 

v2r=av cos θ2R=10R=10010R=10 m
 

Hope this information will clear your doubts about the topic.

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