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Displacement, Velocity, and Acceleration

AP Physics 1· difficulty 4/5

A ball is launched at v0=25 m/sv_0 = 25\text{ m/s} at θ=60°\theta = 60°. What maximum height above the launch point does it reach? Use g=9.8 m/s2g = 9.8\text{ m/s}^2.

  • A

    23.9 m

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  • B

    12.7 m

  • C

    31.9 m

  • D

    15.9 m

Explanation

Vertical component: v0y=25sin⁡60°≈21.65 m/sv_{0y} = 25 \sin 60° \approx 21.65\text{ m/s}. Then H=v0y2/(2g)≈468.7/19.6≈23.9 mH = v_{0y}^2/(2g) \approx 468.7/19.6 \approx 23.9\text{ m}.

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