From the window of a building, a ball is tossed from a height yo above the groun
ID: 1654514 • Letter: F
Question
From the window of a building, a ball is tossed from a height yo above the ground with an initial velocity of 8.90 m/s and angle of 18.0° 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: yo. Assume SI units. Do not substitute numerical values; use variables only.) (b) With the positive x-direction chosen to be out the window, find the x and y-components of the initial velocity. m/s m/s (c) Find the equations for the x- and y-components of the position as functions of time. (Use the following as necessary: Yo and t. Assume SI units.)Explanation / Answer
Here we will use two types of equation of linear motion -
s = ut + 0.5at^2 and s = 0.5(u+v)t
where,
s = displacement, u = initial velocity, v = final velocity,
a = acceleration due to gravity = 9.8 m/s^2
x-component of the initial velocity = 8.9*cos18 m/s
y-component of the initial velocity = 8.8*sin18 m/s
(a) xi = 0, as the building is straight.
yi = s = ut + 0.5at^2 = (8.9sin18)(4) + (9.8/2)(4)^2 = 11.0 + 78.4 = 89.4 m
(b) As calculated above -
x-component of the initial velocity = 8.9*cos18 m/s = 8.46 m/s
y-component of the initial velocity = 8.9*sin18 m/s = 2.75 m/s
(c) x = s = 0.5(u+v)t = 0.5(8.9*cos18+8.9*cos18)t = 8.9*cos18t
Please note that, here u = v because in projectile motion the horizontal velocity is not affected by acceleration from gravity, which acts only vertically.
and
y = s = ut + 0.5at^2 = (8.9*sin18)t + (9.8/2)(t)^2 = (8.9*sin18)t + 4.9*(t)^2
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