Question #7ef68

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Question #7ef68

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Answer 1 · 2016-05-18T19:35:12

$h = 25 \cdot 5 = 125 m e t e r s$ $h=25*5=125 " "meters"$

Explanation:

$let x be displacement in the first second$ $" let x be displacement in the first second"$
$for t=2 \to 3 x$ $"for t=2" rarr 3x$
$for t=3 \to 5 x$ $"for t=3" rarr 5x$

$total= x + 3 x + 5 x = 9 x$ $"total="x+3x+5x=9x$

$thus:$ $"thus:"$

$x, 3 x, 5 x, 7 x, 9 x$ $x,3x,5x,7x,9x$

$total distance= x + 3 x + 5 x + 7 x + 9 x = 25 x$ $"total distance="x+3x+5x+7x+9x=25x$

$x = \frac{1}{2} \cdot g \cdot t^{2} = \frac{1}{2} \cdot 10 \cdot 1 = 5 displacement for the first second$ $x=1/2*g*t^2=1/2*10*1=5" displacement for the first second "$

$h = 25 \cdot 5 = 125 m e t e r s$ $h=25*5=125 " "meters"$

Answer 2 · 2016-05-18T19:53:22

$h = 125, s_{3} = 45, t_{f} = 5$ $h = 125, s_3 = 45, t_f = 5$

Explanation:

The space covered during the fall is given by
$s = h - \frac{1}{2} g \cdot t^{2}$ $s = h-1/2g* t^2$
after $3$ $3$ [s] we have
$e q_{1} \to h - s_{3} = h - \frac{1}{2} g \cdot 3^{2}$ $eq_1->h-s_3 = h-1/2 g* 3^2$
Being $t_{f}$ $t_f$ the impact time we have
$e q_{2} \to 0 = h - \frac{1}{2} g \cdot t_{f}^{2}$ $eq_2->0=h-1/2g*t_f^2$
and at $t_{f} - 1$ $t_f-1$ we have
$e q_{3} \to s_{3} = h - \frac{1}{2} g \cdot {(t_{f} - 1)}_{f}^{2}$ $eq_3->s_3=h-1/2g*(t_f-1)^2$
Solving ${e q_{1}, e q_{2}, e q_{3}}$ ${eq_1,eq_2,eq_3}$ for $h, s_{3}, t_{f}$ $h,s_3,t_f$ we obtain
$h = 125, s_{3} = 45, t_{f} = 5$ $h = 125, s_3 = 45, t_f = 5$

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Question #7ef68

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