Q.A ball is projected with a velocity of 20 m/s at an angle of 45 degrees (g = 10 m/s^2). Using R = v^2 sin(2*theta)/g, its horizontal range is:

Imagine you throw a ball across a park. You want it to land as far away from you as possible. The distance between where you threw it and where it first hits the ground is what we call the range of the projectile. In everyday life, you already have a feel for this: a cricketer hitting a six is trying to maximise the range of the ball; a firefighter aiming a hose at a burning building is trying to control the range of the water jet.

The precise meaning of projectile motion range is the horizontal distance a projectile travels from its launch point to the point where it returns to the same vertical level from which it was launched. "Same vertical level" is key — it means the launch height and the landing height are the same. If you throw a ball from ground level and it lands on ground level, the range is the straight-line horizontal distance between those two points. If you throw it from a cliff into the sea, the range is still the horizontal distance, but the landing height is lower, so the calculation changes — but the core idea of "how far it goes sideways" remains.

Why does range matter? It is the most practical measure of a projectile's effectiveness in sports, engineering, and even warfare. A javelin thrower cares only about range; an artillery shell's range determines whether it can hit a target; a basketball player's shot has a range that must match the distance to the hoop. In physics, range is the outcome of two competing factors: how fast you launch the object (its speed) and the angle at which you launch it. A gentle toss at a low angle gives a short range; a hard throw at a steep angle might go high but not far. The sweet spot — the angle that gives the maximum possible range for a given speed — is a famous result, but you do not need the formula to understand the idea. …