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1- an object is thrown down at 6m/s from the top of a 24m tall building. determine the time it takes the object to reach the ground. remember that it is important that it is grown down.

2- an object is thrown upwards at 20m/s from the ground. what are the 2 times that it reaches 8m?

Respuesta :

onyebuchinnaji

(1) the time for the object to reach the ground is 1.68 s.

(2) The two times that the object reaches 8m is 3.63 s and 0.45 s

The given parameters;

  • initial velocity of the object, u = 6 m/s
  • height of the building, h = 24 m

(1) The time taken for the object to reach the ground is calculated as;

[tex]h = ut + \frac{1}{2} gt^2\\\\24 = 6t + 0.5\times 9.8t^2\\\\24= 6t + 4.9t^2\\\\4.9t^2 + 6t - 24 =0\\\\solve\ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ b = 6, \ c = -24\\\\t = \frac{-b + /- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-6 + /- \ \ \sqrt{(6)^2-4(4.9\times -24)} }{2(4.9)} \\\\t = 1.68 \ s[/tex]

Thus, the time for the object to reach the ground is 1.68 s.

(2)

The given parameters;

  • initial velocity of the object, u = 20 m/s
  • height of the building, h = 8 m

The two times that the object reaches 8m is calculated as;

[tex]h = ut - \frac{1}{2} gt^2\\\\8 = 20t - 0.5\times 9.8t^2\\\\8 = 20t - 4.9t^2\\\\4.9t^2 -20t+ 8 = 0\\\\[/tex]

[tex]solve\ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ b = -20, \ c = 8\\\\t = \frac{-b + /- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-20)\ + /- \ \ \sqrt{(-20)^2-4(4.9\times 8)} }{2(4.9)} \\\\t = 3.63 \ s \ \ or \ 0.45 \ s[/tex]

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