a.) the ball was going up at 0.95 seconds and coming down at 4.58 seconds
b.) the distance of the wall from the plate is 34 m
Given that a ball player hits a home run, and the ball just clears the wall which is 22.0 m high. The ball is hit at an angle of 37.0" with a velocity of 45.0 m/s.
Since the ball is hit at an angle of 37.0 degrees, We need to find the vertical and horizontal component of the velocity.
[tex]U_{y}[/tex] = 45 Sin 37 = 27.08 m/s
[tex]U_{x}[/tex] = 45 cos 37 = 35. 94 m/s
Let us first calculate the maximum height reached.
[tex]V^{2}[/tex] = [tex]U_{y} ^{2}[/tex] - 2gH
At maximum height, V = 0
0 = [tex]27.08^{2}[/tex] - 2 x 9.8H
19.6H = 733.3
H = 733.33/19.6
H = 37.4 m
If the ball is hit from a height of 0.750 m above the ground,
(a) To calculate the time the ball stays in the air until it clears the wall while on its way down, we will use the formula below.
h = [tex]U_{y}[/tex]t - 1/2g[tex]t^{2}[/tex]
Substitute for all the parameters
22 - 0.75 = 27.08t - 0.5 x 9.8 x [tex]t^{2}[/tex]
21.25 = 27.08t - 4.9[tex]t^{2}[/tex]
4.9[tex]t^{2}[/tex] - 27.08t + 21.25
We will use quadratic formula
a = 4.9
b = - 27.08
c = 21.25
t = [tex]\frac{-b+/- \sqrt{b^{2} - 4ac } }{2a}[/tex]
t = [tex]\frac{27.08 +/- \sqrt{27.08^{2} - 4 * 4.9 * x^{2} 21.25 } }{2 * 4.9}[/tex]
t = [tex]\frac{27.08 +/-\sqrt{733.3 - 416.5} }{9.8}[/tex]
t = [tex]\frac{27.08 +/- 17.8}{9.8}[/tex]
t = 44.88/9.8 or 9.28 / 9.8
t = 4.58 s or 0.95 s
This means that the ball was going up at 0.95 seconds and coming down at 4.58 seconds
(b) The distance of the wall from home plate will be the range which is
R = [tex]U_{x}[/tex]t
R = 35.94 x 0.95
R = 34.143 m
Therefore, the distance of the wall from the plate is 34 m approximately
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