An object is thrown vertically from a height of $2 m$ $2 m$ at $2 \frac{m}{s}$ $2 m/s$ . How long will it take for the object to hit the ground?

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An object is thrown vertically from a height of $2 m$ $2 m$ at $2 \frac{m}{s}$ $2 m/s$ . How long will it take for the object to hit the ground?

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Answer 1 · 2017-04-12T17:58:28

0.56518s

Explanation:

For a distance, $d$ $d$ , an initial velocity, $v_{0}$ $v_0$ , an acceleration of $a$ $a$ , and a time of $t$ $t$
$XXX d = v_{0} t + a t^{2}$ $color(white)("XXX")d=v_0t+at^2$

In this case
$XXX d = - 2 m$ $color(white)("XXX")d=-2 " m"$ (2 meters below the height from with it was thrown)If it was dropped and not thrown up there would be no negative sign
$XXX v_{0} = 2 \frac{m}{sec}$ $color(white)("XXX")v_0= 2 " m"/"sec"$
$XXX a = - 9.8 \frac{m}{{sec}^{2}}$ $color(white)("XXX")a=-9.8 " m"/(sec"^2)$ (standard gravity assumed)

So we have
$XXX - 2.0 m = 2.0 t m + (- 9.8 t^{2}) m$ $color(white)("XXX")-2.0" m" = 2.0t " m" + (-9.8t^2)" m"$

$XXX 98 t^{2} - 20 t - 20 = 0$ $color(white)("XXX")98t^2-20t-20=0$

Using the quadratic formula:
$XXX t = \frac{20 \pm \sqrt{20^{2} - 4 (98) (- 20)}}{2 (98)}$ $color(white)("XXX")t=(20+-sqrt(20^2-4(98)(-20)))/(2(98))$

$XXX t = 0.56518 s$ $color(white)("XXX")t = 0.56518s$

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An object is thrown vertically from a height of $2 m$ $2 m$ at $2 \frac{m}{s}$ $2 m/s$ . How long will it take for the object to hit the ground?

1 Answer

Explanation:

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