Physics · Motion In A Plane · NEET
Start with the range formula R = v0^2 sin(2θ)/g. Here v0 and g are fixed, so R depends only on sin(2θ). The sine of any angle is largest (equal to 1) when that angle is 90 degrees. So we need 2θ = 90 degrees, which gives θ = 45 degrees. Putting sin(2θ) = 1 back in gives Rmax = v0^2/g. Nothing else can make sin(2θ) bigger than 1, so 45 degrees is the unique answer.
No, and this is a very common NEET trap. Maximum height is H = v0^2 sin^2(θ)/(2g). Height keeps growing as θ increases and is largest at θ = 90 degrees (straight up). Range is largest at θ = 45 degrees. So maximum height and maximum range happen at different angles. At 45 degrees you get maximum range, but the height there is only half of the range value (H = Rmax/4).
Because sin(2θ) gives the same value for an angle and its complement. For 30 degrees, 2θ = 60 degrees; for 60 degrees, 2θ = 120 degrees; and sin(60) = sin(120). More generally, angles θ and (90 - θ) always produce equal ranges. This is Galileo's rule: angles above and below 45 degrees by the same amount give equal ranges. Only 45 degrees itself gives the single maximum, because its complement (also 45) is the same angle.
At θ = 45 degrees, 2θ = 90 degrees, and sin(90 degrees) = 1. So R = v0^2 sin(2θ)/g becomes R = v0^2 (1)/g = v0^2/g. That is the maximum possible range for a given launch speed v0. If you double the speed, Rmax becomes four times larger, because range depends on v0 squared.
For NEET, we ignore air resistance, so the answer is always 45 degrees. In real life with air drag, the best angle is slightly less than 45 degrees (often around 40 degrees for a thrown ball). But in every NEET problem the standard model has no air resistance, so use 45 degrees for maximum range with full confidence.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
45 degrees. At this angle a projectile with a given speed covers the greatest horizontal distance, giving Rmax = v0^2/g.
Rmax = v0^2/g, obtained by putting θ = 45 degrees (so sin(2θ) = 1) into R = v0^2 sin(2θ)/g.
Yes. They are complementary (they add to 90 degrees), so sin(2θ) is the same for both, and the range is equal. This follows Galileo's complementary-angle rule.
At 45 degrees, H = Rmax/4. So the maximum height is one quarter of the maximum range for that launch.
NEET regularly tests the 45-degree result, the Rmax = v0^2/g formula, and the trap of confusing max range with max height. Knowing these lets you solve projectile MCQs in seconds.
