Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take `g = 10 ms^(-2)`.
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The projection speed is :

Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take g = 10 ms^(-2) . Projection angle with the horizontal is :

Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take g = 10 ms^(-2) . Projection angle with the horizontal is :

Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take g = 10 ms^(-2) . , Maximum height attained from the point of projection is

Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take g = 10 ms^(-2) . , Maximum height attained from the point of projection is

Two graphs of the same projectile motion (in the x - y plane) projected from origin are shown in (Fig. 5 .10). X - axis is along horozontal direction and Y - axis is vertically upwards. Take g = 10 m s^2 Find (i) The y component of initial velocity and (ii) the X component of initial velocity. .

A particle is projected from gound At a height of 0.4 m from the ground, the velocity of a projective in vector form is vecv=(6hati+2hatj)m//s (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is (g=10m//s^(2))

A particle is projected from gound At a height of 0.4 m from the ground, the velocity of a projective in vector form is vecv=(6hati+2hatj)m//s (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is (g=10m//s^(2))

A projectile is thrown from ground at a speed v_(0) at an angle alpha to the horizontal. Consider point of projection as origin, horizontal direction as X axis and vertically upward as Y axis. Let t be the time when the velocity vector of the projectile becomes perpendicular to its position vector. (a) Write a quadratic equation in t. (b) What is the maximum angle a for which the distance of projectile from the point of projection always keeps on increasing? [Hint: Start from the equation you obtained in part (a)]

At a height of 0.4 m from the ground, the velocity of projectile in vector from is vec(u)=(6hat(i)+2hat(j)) (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is : (g = 10 m//s^(2))

A bullet is fired from horizontal ground at some angle passes through the point (3R/4 ,R/4), where R is the range of the bullet. Assume point of the fire to be origin and the bullet moves in x-y plane with x-axis horizontal and y-axis vertically upwards. Then angle of projection is