[" projectile when launched at an angle "45" to une nuritile launched from the origin is given by : "],[" Show that the projection angle "theta_(0)" for a projectile "],[theta_(0)=tan^(-1)[(4H_(m))/(R)]quad " Where "H_(m)=" Maximum height,"R=" Horizontal range "]

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

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

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

Show that the projection angle theta_0 for a projectile launched from the origin is given by theta_0=tan^(-1)((4H)/R) Where the symbols have their usual meaning.

Show that the projection angle thetha_0 for a projectile launched from the origin is given by theta_0 =tan^1(4h_m//R) Where the symbols have their usual meaning.

Shows that the projection angle theta_o for a projectile launched from the origin is given by theta_o=tan^-1((4h_m)/R) where the symbols have their usual meaning.

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.

The range of a projectile when launched at angle theta is same as when launched at angle 2theta . What is the value of theta ?