A ball thrown horizontally at 22.0 m/s from the roof of a building lands 40.0 m from the base of the building. How high is the building?
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Shelly A. asked • 01/06/16
A ball thrown horizontally at 22.0 m/s from the roof of a building lands 40.0 m from the base of the building. How high is the building?
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Sanhita M. answered • 01/06/16
TUTOR 4.7 (11)
Mathematics and Geology
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let's think air is static and not influencing the motion of the ball.
Thus the ball has an initial velocity v0=22.0 m/s and vertical acceleration, g= acceleration due to gravity= 9.8 m/s
It covers horizontal distance, x=40.0 m supposedly in t seconds
Since the ball has been thrown horizontally the motion path of the ball, the projectile, yields an angle of 0 degree with the horizontal axis of references.
The vertical distance, say, y, traveled in time t seconds hence the height of the building.
Hence we have, 40=22tcos0
=> 40= 22t ... since cos0=1
=>t=40/22
y= 22tsin0-1/2(gt2) m
=0-0.5(9.8)(40/22)2 m . since sin0=0
=-(4.9)(1.818)2 m . rounded up to 3 decimal place
=-(4.9)(3.305124) m
=-16.1951076 m
=-16.195 m
We can ignore the negative sign as it denotes the downward motion of the ball.
Therefore, rounding up to three decimal place, the building seems 16.195 meter high.