# When a projectile is launched what is the expression for the range in terms of initial velocity and the angle of projection

justaguide | College Teacher | (Level 2) Distinguished Educator

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The range of a projectile fired with an initial velocity v m/s at an angle `theta` to the horizontal has to be determined in terms of v and theta.

Divide the initial velocity of the projectile into its components. The component of velocity in the horizontal direction is `v*cos theta` . The vertical component is `v*sin theta` . The horizontal component of velocity is not changed. The vertical component changes due to gravitational acceleration. If the time that the projectile is in flight is T,` -v*sin theta = v*sin theta - g*T`

=> `T = (2*v*sin theta)/g`

The horizontal distance traveled by the particle in time T is `v*cos theta*T` = `(v*cos theta)*((2*v*sin theta)/g)`

= `(v* sin(2*theta))/g`

The range of a particle fired at v m/s at an angle `theta` to the horizontal is given by `(v* sin(2*theta))/g`