After going through the hoop, a basketball falls to the floor 3 meters below the hoop. The ball
1 answer:
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Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft
Explanation:
1. use the position (x) equation in parobolic movement to find the time (t)
565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t
t= 3.92 s
2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.
y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] -
y= 148.32 ft
3. finally add the 3 ft that exist between the home plate and the ball
148.32 ft + 3 ft = 151.32
Answer:
Answer:196 Joules
Explanation:
Hello
Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem
the work is the product of a force applied to a body and the displacement of the body in the direction of this force
assuming that the force goes in the same direction of the displacement, that is upwards
W=F*D (work, force,displacement)
the force necessary to move the object will be
Answer:196 Joules
I hope it helps