You Drop Your Cellphone While In The Bathroom In Front Of The Mirror While The Water Is Running Is In The Sink Below. If You Are 0.45meters Above The

July 28, 2022

You drop your cellphone while in the bathroom in front of the mirror while the water is running is in the sink below. If you are 0.45meters above the sink,how long do you have before your cell phone is a gonner.

y = 0.45 m

a = -9.81 m/s^2

$V_{0} = 0$ m/s

t = ?

$(V_{f})^{2} = (V_{0})^{2} + 2ay$

$V_{0} = \sqrt{2ay}$

$V_{0} = \sqrt{2(-9.81 m/s^2)(-0.45 m)}$

$V_{0} =$ 2.971363323 m/s

$y = V_{0}t + \frac{1}{2} at^{2}$

$0.45 m = 2.971363323 m/st + \frac{1}{2} (-9.81 m/s^2)t^{2}$

$0.45 m = 2.971363323 m/st - (4.905 m/s^2)t^2$

$-4.905 m/s^2t^2 + 2.971363323 m/st - 0.45 m = 0$

$(-4.905 m/s^2t^2 + 2.971363323 m/st - 0.45 m = 0) -1$

$4.905 m/s^2t^2 - 2.971363323 m/st + 0.45 m = 0$

Using the quadratic formula: a = 4.905; b = -2.971363323 m/s; c = 0.45

Just get the positive quotient.

t = 0.302891266 s

Answer:

Approximately t = 0.30 s.

(Note: I am not sure with my answer. Sorry.)

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You Drop Your Cellphone While In The Bathroom In Front Of The Mirror While The Water Is Running Is In The Sink Below. If You Are 0.45meters Above The

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