Question #7cc65

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Question #7cc65

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Answer 1 · 2018-01-28T17:51:08

I have already answered this here:

Explanation:

https://socratic.org/questions/an-object-is-projected-vertically-up-from-the-top-of-a-tower-of-height-58-8-m-wi#224886

Answer 2 · 2018-01-28T17:55:52

(2)

Explanation:

Applicable kinematic expression is

$h = u t + \frac{1}{2} g t^{2}$ $h=ut+1/2g t^2$ .....(1)

Lets choose origin of the coordinate system at the point of projection of object. When object reaches ground level, $h = - 58.8 m$ $h=-58.8m$ . taking value of $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$ and inserting values in (1) we get

$- 58.8 = 4.9 t - \frac{1}{2} \times 9.8 \times t^{2}$ $-58.8=4.9t-1/2xx9.8xxt^2$
$\Rightarrow 4.9 t^{2} - 4.9 t - 58.8 = 0$ $=>4.9t^2-4.9t-58.8=0$

Diving both sides with $4.9$ $4.9$ equation becomes

$\Rightarrow t^{2} - t - 12 = 0$ $=>t^2-t-12=0$

The quadratic can be solved by splitting the middle term.

$\Rightarrow t^{2} - 4 t + 3 t - 12 = 0$ $=>t^2-4t+3t-12=0$
$\Rightarrow t (t - 4) + 3 (t - 4) = 0$ $=>t(t-4)+3(t-4)=0$
$\Rightarrow (t - 4) (t + 3) = 0$ $=>(t-4)(t+3)=0$

Setting both factors equal to zero, we get two roots as $t = 4$ $t=4$ and $t = - 3$ $t=-3$ . Ignoring the $- v e$ $-ve$ root as time can not be $- v e$ $-ve$ , we get

$t = 4.0 s$ $t=4.0s$

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Question #7cc65

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