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

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

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,

Prove that sin ^(2) ((pi)/(8) + (theta)/(2)) - sin ^(2) ((pi)/(8) - (theta)/(2)) = sin ""(pi)/(4) sin theta.

If theta in (0, 2pi) and 2sin^2theta - 5sintheta + 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)

Prove that (2 sin theta - sin 2theta)/(2 sin theta + sin 2theta) = tan^2\ theta/2

The minimum and maximum velocities of a projectile are 10 ms^(-1) and 20 ms^(-1) respectively. The horizontal range and maximum height are respectively (g=10 ms^(-2) )

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