Asked by Anonymous
A cannon ball is shot straight upward with a velocity of 57.50 m/s. How high is the cannon ball above the ground 4.30 seconds after it is fired? (Neglect air resistance.)
Answers
Answered by
Henry
h = Vo*t + 0.5g*t^2,
h = 57.5*4.3 + (-4.9)(4.3)^2 = 156.6 m.
h = 57.5*4.3 + (-4.9)(4.3)^2 = 156.6 m.
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