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

Shown 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_(oy)-gt)/(v_(ax))) (b) Show that the projection angle theta_(0) for a projectile lauched from the origin is given by theta_(0)=(tan^(-1)(4h_(m))/(R)) Where the symbols have their usual meaning.

Show that for a projectile the angle between the velocity and the x-axis as function of time is given by theta_(t) = tan ^(-1)( ( v_(0 y)-g t)/(v_(o x))) (b) Shows that the projection angle theta_0 for a projectile launched from the origin is given by theta_(t) = tan ^(-1)( ( 4 h_m)/R) where the symbols have their usual meanings.

Maximum range for a projectile motion is given as R, then height will be

Show that the maximum height reached by a projectile launched at an angle of 45^@ is one quarter of its range.

Find the angle of projection for a projectile motion whose rang R is (n) times the maximum height H .

The position of a projectile launched from the origin at t = 0 is given by s = (40hati+50hatj)m at t=2s . if the projectile was launched at an angle theta from the horizontal, then theta is (take g = 10 ms^(–2)

For a projectile 'R' is range and 'H' is maximum height

A projectile is fired from a horizontal frictionless ground. Coefficient of restitution between the projectile and the ground is e. If T_(1), H_(1), R_(1), v_(1), T_(2), H_(2), R_(2), v_(2) , are time flight, maximum height, range, horizontal velocities in first two collisions, then match the following.

A projectile is fixed on a horizontal ground. Coefficient of restitution between the projectile and the ground is 'e'. If a,b and c be the ration of time of flight [(T_(1))/(T_(2))] , maximum height [(H_(1))/(H_(2))] and horizontal range [(R_(1))/(R_(2))] in first two collisions with the ground, then a, b and c in terms of 'e' will be

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