Show that the maximum height attained by a projectile is one fourth of its maximum horizontal range.

Equations of trajectory of a projectile are x=4 t, y= 3t - 5 t ^(2) , where x and y are in metre and t is in second . Find (i) maximum height attained by the projectille , (ii) horizontal range of the projectile .

Prove that the maximum horizontal range is four times the maximum height attained by a projectile which is fired along the required oblique direction.

The equation of the trajectory of a projectile on a vertical plane is y= ax-bx^2 , where a and b are constants, and x and y respectively are the horizontal distances of the projectile from the point of projection. Find out the maximum height attained by the projectile, and the angle of projection with respect to the horizontal.

The resultant velocity of a projectile at its highest point is sqrt(6/7) times that at half the maximum height. Show that the angle of projection with respect to the horizontal is 30^@ .

A particle is projected with an initial velocity u, making an angle theta with the horizontal. Find the equation of the trajectory of the particle at any instant after the projection. Find expressions for the maximum height gained by the particle and its horizontal range.

A stone is thrown up from the earth's surface with an initial velocity of 28 m*s^(-1) , making and angle 30^@ with the horizontal. Find the maximum height attained and the range of the projectile motion of the stone.

Which one has maximum biomass ?

What is a projectile .Show that the path following by a projectile is a parabolic , when it is projected at an angle theta with the horizontal. Obtain an expression for maximum height.

For angles theta and (90^@-theta) of projection, a projectile has the same horizontal range R. The maximum heights attained are H_1 and H_2 respectively. Then the relation among R,H_1 and H_2 is