a baseball is thrown to a distance of 890 m at an angle of 30 degree above horizontal. what is the initial speed of the baseball
Given:
"R=890\\;\\rm m"
"\\theta=30^\\circ"
The range of projectile is given by
"R=\\frac{v_0^2\\sin 2\\theta}{g}"Hence, the initial velocity
"v_0=\\sqrt{\\frac{gR}{\\sin 2\\theta}}=\\sqrt{\\frac{9.8*890}{\\sin 60^\\circ}}=100\\:\\rm m\/s"
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