A projectile is launched at 30° above horizontal with initial speed 20 m/s from ground level. Taking g = 10 m/s², the maximum height reached is closest to

5.0 m (v{0y} = 10 m/s, h = v{0y}^2 / (2g)) v{0y} = 20\sin 30° = 10 m/s. At max height vy = 0, so h = v{0y}^2/(2g) = 100/20 = 5.0 m.

5.0 m (v{0y} = 10 m/s, h = v{0y}^2 / (2g))

Vectors and Motion in Two Dimensions

AP Physics 1· difficulty 4/5

30°

Explanation

v_{0y} = 20\sin 30° = 10

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