A projectile is projected at an angle (a45) with an initial velocity u.The time t ,at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is:

Question

A projectile is projected at an angle (a>45) with an initial velocity u.The time t ,at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is:

Vikas TU · Answer

let the angle tbe theta, then at any time t, in x – direcn. Vx = ucos(theta) in y-direcn. Vy = usin(theta) – gt Now Vx = Vy (given thhen, usin(theta) – gt = ucos(theta) u(sin(theta) – cos(theta)) = gt t = [u(sin(theta) – cos(theta))]/g theta = a {given} t = [u(sin(a) – cos(a))]/g

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A projectile is projected at an angle (a>45) with an initial velocity u.The time t ,at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is:

A projectile is projected at an angle (a>45) with an initial velocity u.The time t ,at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is:

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A projectile is projected at an angle (a>45) with an initial velocity u.The time t ,at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is:

A projectile is projected at an angle (a>45) with an initial velocity u.The time t ,at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is:

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