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Vectors and Motion in Two Dimensions

AP Physics 1· difficulty 3/5

A ball is launched from ground level at 20 m/s20~\text{m/s} at an angle of 30∘30^\circ above the horizontal. Using g≈10 m/s2g \approx 10~\text{m/s}^2, what is its maximum height above the launch point?

  • A

    10 m10~\text{m}

  • B

    5.0 m5.0~\text{m}

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  • C

    15 m15~\text{m}

  • D

    2.5 m2.5~\text{m}

Explanation

Vertical launch speed: vy0=20sin⁡30∘=10 m/sv_{y0} = 20\sin 30^\circ = 10~\text{m/s}. At the top, vy=0v_y = 0, so 0=vy02−2gh0 = v_{y0}^2 - 2gh gives h=vy02/(2g)=100/20=5.0 mh = v_{y0}^2/(2g) = 100/20 = 5.0~\text{m}.

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