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 9.50 m/s and angle of 22.0° 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: 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?

(e) Find the height from which the ball was thrown.

(f) How long does it take the ball to reach a point 10.0 m below the level of launching?

Explanation / Answer

a)
x cordinate = 0
y cordinate = y0

b)
x cordinate of initial velocity = 9.5 * cos 22 = 8.8 m/s
y cordinate of initial velocity = - 9.5 * sin 22 = 3.6 m/s

c)
Acceleration in x cordinate = 0
so,x cordinate = Vx*t = 8.8*t m

Acceleration in x cordinate = -9.8 m/s^2
so,y cordinate = Vy*t - 0.5*9.8*t^2 = -3.6*t - 4.9*t^2 m

d)
since it atkes 5 s
Horizontal distance = Vx*t = 8.8*5 = 44 m

e)
vertical dispalcement when it reaches bottom is -yo
use:
-yo= -3.6*t - 4.9*t^2
-yo = -3.6*5 - 4.9*5^2
yo = 140.5 m

f)
here displacement = - 10
use:
-10= -3.6*t - 4.9*t^2
3.6t + 4.9t^2 -10 = 0

solving for t we get,
t = 1.12 s and -2.5 s

Since time can't be negative, t = 1.12 s
Answer: 1.12 s

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