Change the launch speed and angle, watch the golf ball fly, and connect its horizontal and vertical motion directly to the SUVAT equations.
The tee shot
Ideal projectile model
Step 1Guided explanation
Ready to examine the launch.
Press Play shot. The explanation will update automatically during ascent, maximum height and descent.
Change any variable and connect the animation to the live measurements and equations.
Compare the blue trajectory with a second launch angle.
Challenge: position the hazards, clear the tree and bunker, then land between the green’s white distance markers.
Right-handed golf double pendulum Green link: shoulders, arms and hands Yellow link: club and clubhead The green link rotates anticlockwise continuously. The yellow club stays 90° behind before releasing much faster through impact, with no pause in the green link. Predicted clubhead speed: 42.0 m s⁻¹
Ready on the tee
Horizontal velocity36.9 m s⁻¹
Initial vertical velocity25.8 m s⁻¹
Time of flight5.26 s
Maximum height33.9 m
Initial height0 m
Horizontal range194 m
Rollout12 m
Total distance206 m
Current vertical velocity25.8 m s⁻¹
Elapsed time0.00 s
Resultant speed45.0 m s⁻¹
Velocity direction35.0°
Time to highest point2.63 s
Horizontal: x = (u cos θ)t
Vertical: y = y₀ + (u sin θ)t − ½gt²
vₓ = u cos θ (constant)
vᵧ = u sin θ − gt
Trajectory: y = y₀ + x tan θ − gx²/(2u²cos²θ)
Speed: v = √(vₓ² + vᵧ²)
Resolve the launch velocity: uₓ = 45 cos 35° = 36.9 m s⁻¹ and uᵧ = 45 sin 35° = 25.8 m s⁻¹
Choose a shot, then play it to see whether it carries the bunker and finishes on the green.
No drag: mass cancels from F = ma, so changing ball mass does not change the trajectory.
A-Level learning milestones
Resolve velocity into horizontal and vertical components
Treat horizontal and vertical motion independently
Explain velocity and acceleration at maximum height
Apply SUVAT with correct signs and substitutions
Interpret displacement–time and velocity–time graphs
Compare complementary launch angles
Evaluate ideal-model assumptions and air resistance
Explain when mass affects projectile motion
What is happening?
The ball’s launch velocity is resolved into independent horizontal and vertical components. With air resistance ignored, there is no horizontal force, so the horizontal velocity remains constant.
Vertically, the ball accelerates downwards at g. Its vertical velocity decreases to zero at the highest point and then becomes negative as the ball falls.
The exam clue
At maximum height, only the vertical velocity is zero. The ball is still moving horizontally.
A-Level Physics checkpoints
The horizontal and vertical motions share the same time, but otherwise they can be analysed separately.
Use Compare mode with complementary angles such as 30° and 60°. In this ideal model they produce the same range, but take different paths.
Checkpoint: at maximum height, which statement is correct?
Model assumptions
This first model ignores air resistance, wind, lift and spin, and assumes that the ball lands at the same vertical level from which it was launched. A real golf ball does not follow a perfect parabola; realistic golf flight belongs in a clearly separated advanced model.