Show that projection angle `theta_(0)` for a projectile launched from the origin is given by:
`theta_(0)=tan^(-1)[(4H_(m))/(R)]` Where `H_(m)=` Maximum height, R=Range

(a) Show that for a projectile the angle between the velocity and the x - axis as a function of time is given by theta(t)="tan"^(-1)((v_(0y)-"gt")/(v_(@)x)) (b) Shows that the projection angle theta_(@) for a projectile launched from the origin given by theta_(0)=tan^(-1)((4h_(m))/(R)) where the symbols have their usual meaning.

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^(-1)sqrt(2) .

If a projectile is fired at an angle theta with the vertical with velocity u, then maximum height attained is given by:-

A ball is projected with speed u at an angle theta to the horizontal. The range R of the projectile is given by R=(u^(2) sin 2theta)/(g) for which value of theta will the range be maximum for a given speed of projection? (Here g= constant)

A rocket starts vertically upward with speed v_(0) . Show that its speed v at height h is given by v_(0)^(2)-v^(2)=(2hg)/(1+h/R) where R is the radius of the earth and g is acceleration due to gravity at earth's suface. Deduce an expression for maximum height reachhed by a rocket fired with speed 0.9 times the escape velocity.

Assertion: In projectile motion, when horizontal range is n times the maximum height, the angle of projection is given by tan theta=(4)/(n) Reason: In the case of horizontal projection the vertical increases with time.

In a projectile motion let t_(OA)=t_(1) and t_(AB)=t_(2) .The horizontal displacement from O to A is R_(1) and from A to B is R_(2) .Maximum height is H and time of flight is T .If air drag is to be considered, then choose the correct alternative(s).

Three particles are projected vertically upward from a point on the surface of the earth with velocities sqrt(2gR//3) surface of the earth. The maximum heights attained are respectively h_(1),h_(2),h_(3) .

Show that the equation of the line passing through the origin and making an angle theta with the line y=mx +c is (y)/( x) = ( mpm tan theta)/(1 pm m tan theta ) .

Two projectiles are projected at angles ( theta) and ((pi)/(2) - theta ) to the horizontal respectively with same speed 20 m//s . One of them rises 10 m higher than the other. Find the angles of projection. (Take g= 10 m//s^(2) )