An object is thrown vertically from a height of $9 m$ $9 m$ 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 $9 m$ $9 m$ 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 · 2016-05-13T05:18:23

time $t = 1.574623134$ $t=1.574623134" "$ seconds if thrown upward
time $t = 1.166459869$ $t=1.166459869" "$ seconds if thrown downward

Explanation:

The formula for free falling bodies

$s = v_{0} \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $s=v_0*t+1/2*g*t^2$
given $s = 9$ $s=9" "$ meters
$v_{0} = 2$ $v_0=2" "$ meter/second

There are two solutions for the problem
(1) If the object is thrown vertically upward
$s = v_{0} \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $s=v_0*t+1/2*g*t^2$
$- 9 = (2) \cdot t + \frac{1}{2} (- 9.8) \cdot t^{2}$ $-9=(2)*t+1/2(-9.8)*t^2$

$4.9 t^{2} - 2 t - 9 = 0$ $4.9t^2-2t-9=0$

Using quadratic formula

$t = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (4.9) (- 9)}}{2 (4.9)}$ $t=(-(-2)+-sqrt((-2)^2-4(4.9)(-9)))/(2(4.9))$

$t_{1} = + 1.574623134$ $t_1=+1.574623134" "$ seconds
$t_{2} = - 1.166459869$ $t_2=-1.166459869" "$ seconds

(2) If the object is thrown vertically downward
$s = v_{0} \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $s=v_0*t+1/2*g*t^2$
$- 9 = (- 2) \cdot t + \frac{1}{2} (- 9.8) \cdot t^{2}$ $-9=(-2)*t+1/2(-9.8)*t^2$

$4.9 t^{2} + 2 t - 9 = 0$ $4.9t^2+2t-9=0$

Using quadratic formula

$t = \frac{- (+ 2) \pm \sqrt{{(+ 2)}^{2} - 4 (4.9) (- 9)}}{2 (4.9)}$ $t=(-(+2)+-sqrt((+2)^2-4(4.9)(-9)))/(2(4.9))$

$t_{1} = + 1.166459869$ $t_1=+1.166459869" "$ seconds
$t_{2} = - 1.574623134$ $t_2=-1.574623134" "$ seconds

God bless....I hope the explanation is useful.

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An object is thrown vertically from a height of $9 m$ $9 m$ 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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An object is thrown vertically from a height of $9 m$ $9 m$ at $2 \frac{m}{s}$ $2 m/s$ . How long will it take for the object to hit the ground?