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 y0 above the ground with an initial velocity of 8.10 m/s and angle of 15.0° 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.)

(b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity.

(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. Let the variable t be measured in seconds.)

(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 Ball = (0, Yo)
Xi = 0
Yi = Yo m

(b)
Initial Horizontal Velocity Vxi = 8.10 * cos(15) m/s = 7.82 m/s
Initial Vertical Velocity = 8.10 * sin(15) m/s = - 2.1 m/s

(c)
X = 7.82 * t

Y = - 2.1* t - 0.5*9.8*t^2
Y = -2.1 t - 4.9 t^2

(d)
Horizontal distance travelled by ball = 7.82 * 4 m
Horizontal distance travelled by ball = 31.3 m

(e)
Y = -2.1 t - 4.9 t^2
Y = -2.1 * 4 - 4.9 * 4^2
Y = - 86.8 m

Height from which ball was thrown = 86.8 m

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