HSC Physics · NSW Year 11 & 12

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Calculation · 3 marks · Module 5

A projectile is launched at 25 m/s at 30° above the horizontal. Calculate its maximum height. (g = 9.8 m/s²)

Your answer

Vertical velocity = 25 sin30° = 12.5 m/s. At max height v_y = 0, so h = v_y² / (2g) = 12.5² / (2 × 9.8) ≈ 7.97 m.

Feedback3 / 3

Correct method and answer with units. Good practice to state that horizontal velocity stays constant and that you resolved the launch velocity into components first.

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Advanced MechanicsElectromagnetismThe Nature of LightFrom the Universe to the Atom
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