From the window of a building, a ball is tossed from a height y_0 above the grou

Question

From the window of a building, a ball is tossed from a height y_0 above the ground with an initial velocity of 7.80 m/s and angle of 16.0 degree below the horizontal. It strikes the ground 5.00 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: y_0.Assume SI units. Do not substitute numerical values; use variables only.) x_i = y_i = (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. V_i, x = m/s v_i, 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: y_0 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

With the origin at ground level directly below the window, the original coordinates of the
ball are (x, y)=

x i = 0, yi = yo

(b)

vox = vo cos theta = 7.8 cos (-16) = 7.49 m/s

voy =vo sin theta = 7.8 sin (-16) =-2.14 m/s

(c)

Apply kinematic equation

x= xo + vox t + 1/2 ax t^2

=0+ ( 7.49) t + 1/2 ( 0) t^2

x= 7.49 t

y = yo+ voy + 1/2 ay t^2

=yo-(2.14) t - 4.9 t^2

(d)

Since the ball hits the ground at t = 5.00 s, the x-coordinate at the landing site is

x landing = 7.49(5) =37.45 m/s

e)

y = 0+(2.14) 5 +4.9 * 5^2

=133.2 m

(f)

When the ball has a vertical displacement of y = 10 0 . m, it will be moving downward
with a velocity given

vy = - sqrt voy^2 + 2ay ( del y) = - sqrt ( -2.14)+ 2 ( -9.8)(-10) = -13.92 m/s

t= vy-voy/ay = -13.92-(-2.14)/-9.8 =1.2 s

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