A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat
er. Find the true velocity of the swimmer as a vector. (Assume that the i vector points east, and the j vector points north.)
1 answer:
Answer:
<u>velocity of swimmer relative to ground = 3 i -5 j</u>
Explanation:
- To cross a river the swimmer swims relative to river in perpendicular direction.
Velocity of river = -5 j (south)
Velocity of swimmer relative to river = 3 i(north)
So
<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>Velocity of swimmer relative to ground = 3 i -5 j
So magnitude of total velocity is =
=
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Explanation:
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The launch velocity of the marble launcher is 34.65 m/s
Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).
The time for landing can be calculated by the second equation of motion formula:
h = ut + g
Let u = 0
0.2 = 0×t + × 9.8 ×
=
= 0.04
t = 0.2s
Now, the launch velocity of the marble launcher can be calculated by:
Speed = Distance / Time
Speed =
Speed =
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Therefore, the launch velocity of the marble launcher is 34.65 m/s
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