rom the window of a building, a ball is tossed from a height y0 above the ground

Question

rom 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 16.0° below the horizontal. It strikes the ground 6.00 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 (d) How far horizontally from the base of the building does the ball strike the ground? m (e) Find the height from which the ball was thrown. m (f) How long does it take the ball to reach a point 10.0 m below the level of launching? s

Explanation / Answer

a) initial coordinates of the ball is (0,y0)

b) vi,x = 9.5*cos(16) = 9.13 m/sec

vi,y = -9.5*sin(16) = -2.62 m/sec

C) x = Vi,x*t = 9.13*t m

-y = (vi,y*t)+(0.5*g*t^2)

-y= (-2.62*t)+(0.5*9.8*t^2)

-y = (-2.62*t)+(4.9*t^2)

y = (2.62*t)-(4.9*t^2)

d) x = vi,x*t = 9.13*6 = 54.78 m

e) using

y = (2.62*6)-(4.9*6^2)

y = -160.68 m

f) y = (2.62*t)-(4.9*t^2)

-10 = (2.62*t)-(4.9*t^2)

t = 1.72 sec

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