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seraphim [82]
1 year ago
6

1) A ball is thrown upwards with a velocity of 31 m/s. What is its height after 1.7 seconds in the air?

Physics
1 answer:
adoni [48]1 year ago
3 0

When an object is thrown upward, the acceleration due to gravity will be negative. With the use of formula, its height after 1.7 seconds in the air is 38.5 meters

<h3>Motion Under Gravity</h3>

Motion under gravity is also known as vertical motion. When an object is thrown upward, the acceleration due to gravity will be negative.

Given that a ball is thrown upwards with a velocity of 31 m/s. What is its height after 1.7 seconds in the air?

The parameters to be considered are

  • Initial velocity u = 31 m/s
  • height h = ?
  • Time t = 1.7 s

The formula to use is

h = ut - 1/2gt²

h = 31 × 1.7 - 1/2 × 9.8 × 1.7²

h = 52.7 - 14.161

h = 38.539 m

Therefore, its height after 1.7 seconds in the air is 38.5 meters approximately.

Learn more about Vertical Motion here: brainly.com/question/4441382

#SPJ1

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A strong weightless rope has a mass, m, hanging from the middle of it. The tension force on each rope is 25 N, and the rope droo
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By using Lami's theorem, Mass m = 1.75 kg approximately

Given that a strong weightless rope has a mass, m, hanging from the middle of it. If the tension force on each rope is 25 N, and the rope droops at an angle of 20.0 degrees to the horizontal.

By using Lami's theorem, we can get how much mass is hanging from the rope.

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Therefore, 1.75 kg mass is hanging from the rope.

Learn more about resolution of forces here: brainly.com/question/1858958

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