Part A A ball is thrown horizontally from the top of a 69.0-m building and lands

Question

Part A A ball is thrown horizontally from the top of a 69.0-m building and lands 49.0 m from the base of the building. Ignore air resistance. How long is the ball in the air? Submit Answer Tries 0/10 Part B What must have been the initial horizontal component of the velocity? Submit Answer Tries 0/10 Part C What is the vertical component of the velocity just before the bal hits the ground? (Take vertically up to be positive.) Submit Answer Tries 0/10 Part D What is the velocity of the ball just before it hits the ground? Magnitude angle (below the x-axis) Submit Answer Tries 0/10

Explanation / Answer

A] using second equation of motion in vertical, h = 0.5gt^2

t = sqrt(2h/g) = sqrt(2*69/9.8) = 3.75 s

B] initial horizontal velocity = d/t = 49/3.75 = 13.06 m/s

C] v = 0 - gt = -9.8*3.75 = -36.8 m/s

D] magnitude = sqrt(13.06^2 + 36.8^2) = 39.0 m/s

angle = arctan(36.8/13.06) = 70.46 degree

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