1 Answer | Add Yours
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`
We’ve answered 319,661 questions. We can answer yours, too.Ask a question