Mar 30, 2016 · The value of x for which speed of particle is maximum is [Consider gravity free space.] (A) d 2 (B) 2 d (C) d (D) 2d x 2d q q Space for Rough work 9. 9 27. A particle is thrown at an angle with horizontal along the inclined plane of inclination .
a particle is projected at angle tita to the horizontal with initial velocity u.if the horizontal distance and maximum height reached are 20m and 10m respectively. find the value of tita and u . Physics. A ball projected with an initial velocity u at an angle tita to the horizontal cover a horizontal range of 26m after 5sec. Cal.

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The radius of curvature of the trajectory when it makes an angle ((θ/2) ) with the horizontal is (g - acceleration due to gravity) Q. A body is projected with a speed 'u' at an angle $'\theta'$ with the horizontal.
The Superintendent of Documents of the U.S. Government Publishing Office requests that any reprinted edition clearly be labeled as a copy of the authentic work with a new ISBN. e:\seals\gpologo2.eps. U . S . G O V E R N M E N T P U B L I S H I N G O F F I C E

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If I have a desired 2D trajectory where I know the desired speed of my ball and desired path, and a ball that starts at (x=0,y=0), how can I calculate the required force / point at which the force ...
At the starting point, let the radius of curvature is R. Then, From the equation of centripetal acceleration, u 2 R = g cos θ ⇒ R = u 2 g cos θ When the particle is at its highest point, its velocity is u cos θ. Therefore, the radius of curvature is, u cos θ 2 R m i n = g ⇒ R m i n = u 2 c o s 2 θ g. Hope this information will clear ...

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Question From – Cengage BM Sharma MECHANICS 1 KINEMATICS-2 JEE Main, JEE Advanced, NEET, KVPY, AIIMS, CBSE, RBSE, UP, MP, BIHAR BOARD QUESTION TEXT:- A parti...
20. A particle is thrown with a velocity u at an angle θ from the horizontal. Another particle is thrown with the same velocity at an angle a from the vertical. The ratio of times of flight of the two particles will be 1) Tan 2 θ: 1 2) Cot 2 θ: 1 3) Tan θ: 1 4) Cot θ: 1 21. The horizontal and vertical displacement of a projectile are given ...

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At the later time, the projectile still has this horizontal component to its velocity, v x (since the horizontal speed of a projectile is constant). If we know that, and are given the new angle B, solving for the new overall speed at that time means solving for the hypotenuse of a right triangle where v x (=u*cos(A)) is the side adjacent to angle B. If you can do that, you can determine the ...

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