If for a projectile , the velocity of projection is u , range R, time of flight T and maximum height attained H show that, `g^2T^4-4T^2u^2+4R^2=0` and `16gH^2-8u^2H+gR^2=0`.

Maximum height of a projectile is H an range of projection is R when the projectile is projected with the velocity u. Show that R^2 = 16H (u^2/(2g)-H) .

An object is projected vertically upwards with a velocity u from the surface of the earth. Show that the maximum height reached by the object is h=(u^2R)/(2gR-u^2) where R is the radius of the earth. Calculate escape velocity from it.

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

The maximum height attained and the range of a projectile are H and R respectively. If projected with an initial velocity of u, show that, R^2=16H((u^2)/(2g)-H).

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 .

A particle is projected vertically upwards and is at a height h after t_(1) seconds and again after t_(2) seconds then

A train of mass M is in motion. When its velocity is u_1 and acceleration is a_1 , resistive force against its speed is R_1 . Also when velocity is u_2 and acceleration is a_2 , resistive forced is R_2 . If the engine works at a fixed rate P, show that P(u_2-u_1)=u_1u_2(R_1-R_2)+Mu_1u_2(a_1-a_2) .