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the direction of motion of a projectile at a certain instant is inclined at an angle A to the horizon. after t seconds it is inclined at an angle b. find the horizontal component of velocity in terms of g, t, A,, and B.

the direction of motion of a projectile at a certain instant is inclined at an angle A to the horizon. after t seconds it is inclined at an angle b. find the horizontal component of velocity in terms of g, t, A,, and B.

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2 Answers

Shobhit Varshney IIT Roorkee
askIITians Faculty 33 Points
9 years ago
Hi,

Assume where the angle is A, velocity = u
where the angle is B, velocity = v

So, u cosA = v cosB,v = u cosA / cosB

and v sinB = u sinA – gt,
Put v = u cosA / cosB in the above equation
Solve it to get u and then find horizontal velocity by u cosA.

thanks
Kashyap Sindhav
11 Points
5 years ago
Let,The velocity of particle at angle A with horizontal along x be v1 and along y be v2.And it`s velocity at B with horizontal along x be v1 and along y be v3.Now,tanA = v2/v1. And. tanB = v3/v1(Since velocity along x remains constant)v1tanA = v2. And. v1tanB = v3Also, v3 = v2 - gt v1tanB = v1tanA - gtgt = v1tanA - v1tanBgt = v1(tanA - tanB)v1 = gt/(tanA - tanB)

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