Show that the maximum height and range of a projectile are `(U^(2) sin^(2) theta)/(2g) `and `(U^(2) sin 2 theta )/(g)` respectively where the terms have their ragular meanings .

Show that maximum range =(u^2)/(g) .

A particle is projected from the ground with velocity u at angle theta with horizontal. The horizontal range, maximum height and time of flight are R,H and T respectively. They are given by, R = (u^(2)sin 2 theta)/(g),H = (u^(2)sin^(2)theta)/(2g) and T = (2usin theta)/(g) Now keeping u as fixed, theta is varied from 30^(@) to 60^(@) . Then,

A particle is projected from the ground with velocity u at angle theta with horizontal. The horizontal range, maximum height and time of flight are R,H and T respectively. They are given by, R = (u^(2)sin 2 theta)/(g),H = (u^(2)sin^(2)theta)/(2g) and T = (2usin theta)/(g) Now keeping u as fixed, theta is varied from 30^(@) to 60^(@) . Then,

A particle is projected from the ground with velocity u at angle theta with horizontal. The horizontal rnage, maximum hieght and time of flight are R,H and T respectively. They are given by, R = (u^(2)sin 2 theta)/(g),H = (u^(2)sin^(2)theta)/(2g) and T = (2usin theta)/(g) Now keeping u as fixed, theta is varied from 30^(@) to 60^(@) . Then,

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).

If theta in(0,2 pi) and 2sin^(2)theta-5sin theta+2>0 then the range of theta is

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 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)