A particle is projected from the origin in such a way that it passes through a given point P (a, b} What is the minimum requrired speed to do so?

A particle is projected from origin with speed u . Find the minimum value of u if the particle passes through a poiny P (a, sqrt(3)a) .

A particle is projected with an angle of projection theta to the horizontal line passing through the points (P,Q) and (Q,P) referred to horizontal and vertical axes (can be treated as X-axis and Y-axis respectively). The angle of projection can be given by

A particle is projected from a point O with an initial speed of 30 ms^(-1) to pass through a point which is 40 m from O horizontally and 10 m above O. There are two angles of projection for which this is possible. These angles are alpha and beta . The value of |tan (alpha+beta)| is

A ball is projected vertically up such that it passes thorugh a fixes point after a time t_(0) and t_(2) respectively. Find a. The height at which the point is located with respect to the point of projeciton b. The speed of projection of the ball. c. The velocity the ball at the time of passing through point P . d. (i) The maximum height reached by the balll relative to the point of projection A (ii) maximum height reached by the ball relative to point P under consideration. e. The average speed and average velocity of the ball during the motion from A to P for the time t_(1) and t_(2) respectively. .

With what minimum speed must a particle be projected from origin so that it is able to pass through a given point (30m, 40m) ? Take g=10m//s^(2)

Equations of the line passing through the origin and the point (a,b,c) are

A particle is projected from point A perpendicular to inclined plane with a velocity 50m/s as shown in the figure. Particle strikes a vertical plane perpendicularly at point B. Time taken by particle in going from point A to point B is

A particle projected atv a definite angle prop to the horizontal passes through points (a,b) and (b,a) , referred to horizontal and vertical axes through the points of projection. Show that : (a) The horizontal range R = (a^2 + ab + b^2)/(a + b) . (b) The angle of projection prop is given by tan^-1 [(a^2 + ab + b^2)/(ab)] .

A particle is projected from the ground with an initial speed of v at an angle theta with horizontal. The average velocity of the particle between its point of projection and highest point of trajectroy is :