Projectile motion is a favourite JEE and NEET physics topic. Enter the launch speed and angle to get the horizontal range, the maximum height and the time the projectile stays in the air.
Enter the launch speed and angle.
Keep g = 9.8 m/s² for Earth.
Read the range, height and time.
Range: R = u² sin 2θ ÷ g
Max height: H = u² sin² θ ÷ 2g
Time: T = 2u sin θ ÷ g
A projectile is anything thrown or launched that then moves only under gravity, such as a cricket ball, a javelin or water from a hose. This calculator takes the launch speed, the launch angle above the horizontal and the value of g, and returns how far the projectile lands (the range), how high it rises and how long it stays in the air.
It assumes the projectile starts and lands at the same level and ignores air resistance, the standard model in NCERT Class 11 and in JEE and NEET problems. Within that model the answers are exact, and they show the key idea of projectile motion: the horizontal and vertical motions happen at the same time but do not affect each other.
1. Split the launch velocity into components: uₓ = u cos θ horizontally and u_y = u sin θ vertically.
2. Find the time of flight from the vertical motion. The body returns to launch height when u_y t − ½gt² = 0, so T = 2u sin θ ÷ g.
3. Find the maximum height, reached when the vertical velocity is zero: H = (u sin θ)² ÷ 2g.
4. Find the range from the horizontal motion, which has constant velocity: R = uₓ × T = u² sin 2θ ÷ g.
5. Check that the units are consistent: speed in m/s, g in m/s², so H and R come out in metres and T in seconds.
Gravity acts only downwards, so it has no horizontal component. The horizontal velocity u cos θ therefore stays constant for the whole flight, while the vertical velocity starts at u sin θ and falls by g every second, exactly like a ball thrown straight up. Combining x = (u cos θ)t with y = (u sin θ)t − ½gt² and eliminating t gives y = x tan θ − gx² ÷ (2u² cos² θ), which is the equation of a parabola. That is why every projectile path, without air drag, is a parabola.
The range formula R = u² sin 2θ ÷ g depends on sin 2θ, which is largest, equal to 1, when 2θ = 90°, that is θ = 45°. At that angle R = u² ÷ g. Angles equally above and below 45°, such as 30° and 60°, give the same sin 2θ and so the same range, although the steeper throw goes higher and stays up longer. Maximum height keeps increasing with angle and is greatest for a vertical launch, where the range is zero.
Real projectiles feel air resistance, which grows with speed and shortens both the range and the height. For a fast, light object like a shuttlecock the effect is huge; for a heavy shot put it is small. Drag also makes the best angle a little less than 45°. Launching from a height, like a ball thrown from a building, changes the landing time, so the simple formulas no longer apply. The calculator is exact for same-level launches in a vacuum.
Kavya, a school athlete, releases a javelin at 24 m/s at an angle of 38° to the ground, and we ignore air resistance and release height.
Split the velocity: uₓ = 24 cos 38° = 18.9123 m/s, u_y = 24 sin 38° = 14.7759 m/s
Time of flight: T = 2u_y ÷ g = 2 × 14.7759 ÷ 9.8 = 3.0155 s
Maximum height: H = u_y² ÷ 2g = 11.1391 m
Range: R = uₓ × T = 18.9123 × 3.0155 = 57.0296 m
Answer: Range 57.0296 m; Maximum height 11.1391 m; Time of flight 3.0155 s
Having a calculator set to radians while working with angles in degrees when doing the sine and cosine by hand.
Using the same-level formulas for a ball thrown horizontally from a roof or a cliff.
Forgetting that at maximum height the vertical velocity is zero but the horizontal velocity is not.
Measuring the angle from the vertical instead of the horizontal.
Expecting real sports throws to match the ideal range, when air resistance can cut it a lot.
Solving Class 11, JEE and NEET projectile numericals.
Understanding launch angles in cricket, football, long jump and javelin.
Designing water fountain jets and garden sprinkler throw distances.
Comparing how the same throw would travel on the Moon or Mars by changing g.
Checking results from a school lab projectile launcher.
Which angle gives the maximum range?
45°, when launched and landing at the same height without air resistance.
Do 30° and 60° give the same range?
Yes, complementary angles give equal ranges.