Calculate range, maximum height, and time of flight with a visual trajectory graph. Perfect for GCSE and A-Level Physics.
Ground-level launch at angle θ
Quick Examples
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Projectile motion describes the movement of an object launched into the air, affected only by gravity (and ignoring air resistance). The resulting path is a parabola.
The key insight is that horizontal and vertical motions are independent:
Avoid these frequent errors when solving projectile motion questions in GCSE and A-Level Physics exams:
Many students try to use v₀ directly in equations without splitting it into horizontal and vertical components. The initial speed v₀ is NOT the horizontal or vertical velocity alone.
Always start by calculating vₓ = v₀cos(θ) and vy = v₀sin(θ) before anything else.
Horizontal and vertical motions are independent. Horizontal velocity is constant, while vertical velocity changes due to gravity. Students often apply gravity to horizontal motion.
Write horizontal and vertical equations separately. Only gravity affects the vertical direction.
Scientific calculators can be in degree or radian mode. Using the wrong mode gives completely wrong answers for sin, cos, and tan.
Check your calculator mode. For projectile problems, angles are usually given in degrees. Our calculator handles the conversion automatically.
The formula R = v₀²sin(2θ)/g only works for ground-to-ground projectiles. When launched from a height, you need to solve a quadratic equation for the landing time.
For non-zero launch height, use the full equation y = h + vy·t - ½gt² = 0 and solve the quadratic for t.
On flat ground, complementary angles (e.g., 30° and 60°) give the same range. Students often only find one angle.
If θ₁ is a solution, then θ₂ = 90° - θ₁ also gives the same range (unless θ₁ = 45°).
Practice with these GCSE and A-Level style projectile motion questions:
A football is kicked at 20 m/s at an angle of 45° to the horizontal. Calculate the range and maximum height. (Take g = 9.81 m/s²)
Given: v₀ = 20 m/s, θ = 45°, g = 9.81 m/s²
Range: R = v₀²sin(2θ)/g = 20²×sin(90°)/9.81
R = 400 × 1 / 9.81
R = 40.77 m
Max height: H = v₀²sin²(θ)/(2g) = 400×sin²(45°)/(2×9.81)
H = 400 × 0.5 / 19.62
H = 10.19 m
A ball is thrown from the top of a 50 m cliff at 10 m/s at 30° above horizontal. Find the time to hit the ground and the range. (Take g = 9.81 m/s²)
vₓ = 10cos(30°) = 8.66 m/s
vy = 10sin(30°) = 5 m/s
y = 50 + 5t - ½(9.81)t² = 0
4.905t² - 5t - 50 = 0
t = (5 + √(25 + 981))/9.81 = (5 + √1006)/9.81
T = 3.72 s
R = vₓ × T = 8.66 × 3.72
R = 32.2 m
A ball rolls off a table 1.2 m high at 3 m/s. How far from the base of the table does it land?
Time to fall: T = √(2h/g) = √(2×1.2/9.81)
T = 0.495 s
Range: R = v₀ × T = 3 × 0.495
R = 1.48 m
A ball is launched at 20 m/s and needs to reach a target 30 m away on flat ground. At what angle(s) should it be launched?
sin(2θ) = Rg/v₀² = 30×9.81/400 = 0.7358
2θ = arcsin(0.7358) = 47.38°
θ₁ = 23.69° (low trajectory)
θ₂ = 66.31° (high trajectory)
Projectile motion is the curved path of an object launched into the air under the influence of gravity only (no air resistance). The horizontal velocity stays constant while the vertical velocity changes due to gravity.
On flat ground, 45° gives the maximum range. This is because sin(2×45°) = sin(90°) = 1, which maximises the range formula R = v₀²sin(2θ)/g.
Complementary angles (e.g., 30° and 60°) give the same range because sin(2θ) has the same value for both. The low angle gives a fast, flat trajectory; the high angle gives a slower, higher arc.
Launching from a height increases both range and time of flight. The landing time must be found using a quadratic equation because the simple T = 2v₀sin(θ)/g formula only works for ground-level launches.
Horizontal projection is when an object is launched horizontally (θ = 0°) from a height. It has no initial vertical velocity, so it falls under gravity while travelling horizontally at constant speed.
Maximum height occurs when vertical velocity equals zero. Use H = v₀²sin²(θ)/(2g) for ground-level launches, or add the launch height h for elevated launches.
No. Without air resistance, gravity only acts vertically. Horizontal velocity remains constant throughout the entire flight. This is why we treat horizontal and vertical motion independently.
Yes! It covers ground-level launch, launch from height, horizontal projection, and finding launch angles with step-by-step solutions. The Learn Mode guides you through each problem interactively.
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