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a particle is thrown with a speed of u at an angle of alpha with horizontal when a particle makes an angle beta with horizontal then it speed will be

As the horizontal component of velocity remains constant.
Ucos(alpha)=Vcos(beta)
=> V=Ucos(alpha) sec(beta)
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In this case: 

Horizontal velocity remains same = u cos(alpha)

Let particle velocity at time when it makes angle beta be v m/s.

Then,

Vertical velocity = v cos(beta)

horizontal velocity = vertical velocity
u cos(alpha) = v cos(beta)
v = u cos(alpha)/cos(beta)
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Sorry 

it's 

horizontal velocity = horizontal velocity
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It's the easier way u can solve this problem. Best if luck 👍👍👍👍👍 🙂🙂🙂__---ABHIGYAN---__🙂🙂

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capillary a and capillary B of radius R and 2r a dipped in a liquid potential energy of the liquid that rises in capillary capillary p r u n u p respectively UEFA u b is
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