In order to make a ramp that is 3ft high and covers4ft of ground how long must the ramp be
1 answer:
This seems like a right angle triangle situation
the height is 3ft
the base is 4ft
so we're looking for the hypothenus
so we can use the Pythagoras Theorem
which is
the square of the hypothenus is equal to the sum of the square of the two other sides
hyp^2 is equal to opp^2+ adj^2
hyp^2 equals to 3^2 + 4^2
9+16
25
hyp^2 equals to 25
equals to 5
the ramp has to be 5ft long
the height is 3ft
the base is 4ft
so we're looking for the hypothenus
so we can use the Pythagoras Theorem
which is
the square of the hypothenus is equal to the sum of the square of the two other sides
hyp^2 is equal to opp^2+ adj^2
hyp^2 equals to 3^2 + 4^2
9+16
25
hyp^2 equals to 25
equals to 5
the ramp has to be 5ft long
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Answer:
31m/s
23.17°
Explanation:
Given the following :
From the diagram attached :
AB = 15m/s, BC = 18m/s, AC = a = resultant
Resolving Velocity into both vertical and horizontal component.
Kindly see attached picture for detailed explanation.
