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Vectors and Motion in Two Dimensions

AP Physics 1· difficulty 4/5

A projectile is launched at 30 m/s30\,\text{m/s} at 53°53° above horizontal. Use g=10 m/s2g = 10\,\text{m/s}^2, sin⁡53°=0.8\sin 53° = 0.8, cos⁡53°=0.6\cos 53° = 0.6. What is the time at which the projectile returns to its launch height, and the horizontal distance it traveled?

  • A

    t=3.0 st = 3.0\,\text{s}, x=54.0 mx = 54.0\,\text{m}

  • B

    t=4.8 st = 4.8\,\text{s}, x=86.4 mx = 86.4\,\text{m}

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  • C

    t=2.4 st = 2.4\,\text{s}, x=43.2 mx = 43.2\,\text{m}

  • D

    t=6.0 st = 6.0\,\text{s}, x=108.0 mx = 108.0\,\text{m}

Explanation

Vertical velocity component is vy=24 m/sv_y = 24\,\text{m/s}. Time to return to launch height is t=2vy/g=4.8 st = 2v_y/g = 4.8\,\text{s}. Horizontal velocity vx=18 m/sv_x = 18\,\text{m/s} gives x=vxt=86.4 mx = v_x t = 86.4\,\text{m}.

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