1 Answer | Add Yours
When a projectile is fired towards a target, it experiences an acceleration in a direction that is vertically downwards as it moves towards the target. The magnitude of the acceleration is 9.8 m/s^2. This decreases the height of the projectile. To compensate for this the shooter has to aim at a point slightly above the target.
To derive the relation between the height above the target and the vertical distance of the target it is assumed that the projectile is fired in a horizontal direction and the point from which it is fired is raised higher to compensate for the downward motion. This is being done as another way of dealing with the downward acceleration is to change the angle at which the projectile is launched.
If the shooter is at a distance of x meters from the target, the time taken by the projectile to reach the target if it is shot at a speed V is x/V. In this time it moves lower by a distance `(1/2)*9.8*(x/V)^2` . Therefore the projectile should be aimed at a point `y = 4.9*x^2/V^2` m above the target to strike it.
We’ve answered 319,846 questions. We can answer yours, too.Ask a question