From the window of a building, a ball is tossed from a height y0 above the groun

Question

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 9.50 m/s and angle of 23.0° below the horizontal. It strikes the ground 4.00 s later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (Use the following as necessary: y0. Assume SI units. Do not substitute numerical values; use variables only.) xi = . yi = (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. vi, x = m/s vi, y = m/s (c) Find the equations for the x- and y-components of the position as functions of time. (Use the following as necessary: y0 and t. Assume SI units.) x = m y = m

Explanation / Answer

a) xi = 0 and yi = y0

b) v_x = 9.5cos23 = 8.744 m/s

V_y = -9.5sin(23) = -3.711 m/s

c) x = v_x*t = 8.744t m

y = vy - 0.5*a*t^2 = y0 -3.711t - 4.9t^2 Answer

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