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A ball is kicked straight up into the air from a height of 48ft with an initial velocity of 88 ft/s. After how many seconds does the ball hit the ground?

Sagot :

Cricetus

The time taken will be "0.5 seconds".

Newton's Law:

According to the question,

  • Initial velocity, u = 88 ft/s
  • Height, h = 48 ft
  • Acceleration due to gravity, g = 32 ft/s²

By using Newton's equation,

→                     [tex]s = ut+\frac{1}{2}gt^2[/tex]

Substituting the values,

                    [tex]48=88t+0.5\times 32\times t^2[/tex]

[tex]16t^2+886-48=0[/tex]

                       [tex]t = 0.5 \ seconds[/tex]

Thus the above solution is right.

Find out more information about velocity here:

https://brainly.com/question/21320166

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