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iris [78.8K]
3 years ago
13

A ball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resista

nce can be ignored, how high does it go?\
Physics
1 answer:
solniwko [45]3 years ago
3 0
Convert the given in SI units.

         (44 ft/sec)(1 m/ 3.28 ft) = 13.41 m/sec

The distance traveled and the initial velocity can be related through the equation,
  
   d = (Vf)² - (Vi)²/ 2a

where d is the distance, Vf is the final velocity, Vi is the initial velocity, a is the acceleration due to gravity. Substituting the known values from the given above,

    d = ((0 m/s)² - (13.41 m/s)²)/ 2(-9.8 m/s²)

The value of d from the equation,

    d = 9.17 meters

Convert this to feet,

    d = (9.17 m)(3.28 ft / 1 m) = 30 ft

Answer: 30 ft
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time of collision is

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Explanation:

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Now the height attained by the object in the same time is given as

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A 5 N force pushes on the right side of a box. At the same time, a 10 N force pushes on the left side of the box. What happens t
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