WEBVTT

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Here is the problem: A projectile is launched at 12 m/s at 35° from a height of 0 m. Find its time of flight, range and maximum height.

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Treat horizontal and vertical motion separately: constant velocity horizontally, constant downward acceleration g vertically, ignoring air resistance.

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Split the launch velocity into components.

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Time to the highest point.

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Maximum height.

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Time of flight (land at height 0).

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Horizontal range.

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Impact velocity.

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So the answer is Time of flight approximately 1.4032451 s; range approximately 13.793653 m; maximum height approximately 2.414605 m; impact speed approximately 12 m/s.

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Without air resistance the horizontal velocity never changes, while the vertical velocity falls by g every second. The flight ends when the height returns to zero.

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Energy check: v squared at impact equals 144 and v₀ squared plus 2gh₀ equals 144 (equal).

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Level-ground formula: v₀ squared sin 2theta / g equals 13.793653 m.