From the window of a building, a ball is tossed from a height y0 above the groun
ID: 582713 • Letter: F
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 (d) How far horizontally from the base of the building does the ball strike the ground? (e) Find the height from which the ball was thrown. 14.9 m (f) How long does it take the ball to reach a point 10.0 m below the level of launching? 1.10
Explanation / Answer
As per the chegg answer board rules only four questions or sub question should contain in the each question. so i am following the rules of chegg
a) If u take coordinate system
Xi = 9.5 x cos(-23o) = 8.745 m/s.
Yi = 9.5 x sin(-23o) = -3.712 m/s.
d) horizontal component of velocity is 9.5 x cos (23o) = 8.745m/s.
after four seconds the horizontal distance is 4 x 8.745 = 34.98m from the base of the building.
e)
vertical component of the velocity is 9.5 x sin (23o) = 3.712m/s towards down
Height, h = 4 x 3.712 + 0.5 x 9.8 x 42
h = 14.848 + 78.4 = 93.248 m
f)
10m = 3.712 x t + (0.5 x 9.8 x t2)
4.9t2+3.712t-10 =0
solve the quadratic equation. you will get the time which it take the ball reach
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