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

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 .

Solve: 2sin^(2)theta+sin^(2)2theta=2

Solve: 2sin^(2)theta+sin^(2)2theta=2

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,

sin^(3)theta + sin theta - sin theta cos^(2)theta =

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)

The maximum value of the sum of the intercepts made by any tangent to the curve (a sin^(2)theta, 2a sin theta) with the axes is

Prove that sin^(4)theta+sin^(2)thetacos^(2)theta=sin^(2)theta

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

sin2theta+(1)/(3)sin^(3)2theta+(1)/(5)sin^(5)2theta+....=