1010 Time to wait in dropping a stone | Rectilinear Translation | Engineering Mechanics Review at MATHalino

Problem 1010
A stone is thrown vertically up from the ground with a velocity of 300 ft per sec (91.44 m/s). How long must one wait before dropping a second stone from the top of a 600-ft (182.88-m) tower if the two stones are to pass each other 200 ft (60.96 m) from the top of the tower?

English System Solution

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Stone from the ground:

$s = v_it + \frac{1}{2}at^2$

$h_1 = v_{i1}t - \frac{1}{2}gt^2$

$600 - 200 = 300t - \frac{1}{2}(32.2)t^2$

$16.1t^2 - 300t + 400 = 0$

$t = 17.19 \, \text{ sec and } \, 1.44 \, \text{ sec}$

Stone from the top of the tower:

Let t₂ = time to wait before dropping the second stone
$h = \frac{1}{2}g(t - t_2)^2$

With t = 17.19 sec
$200 = \frac{1}{2}(32.2)(17.19 - t_2)^2$

$t_2 = 13.67 \, \text{ sec}$

With t = 1.44 sec
$200 = \frac{1}{2}(32.2)(1.44 - t_2)^2$

$t_2 = -2.08 \, \text{ sec}$ (meaningless)

Use $t_2 = 13.67 \, \text{ sec}$ answer

System International Solution

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Stone from the ground:

$s = v_it + \frac{1}{2}at^2$

$h_1 = v_{i1}t - \frac{1}{2}gt^2$

$182.88 - 60.96 = 91.44t - \frac{1}{2}(9.81)t^2$

$4.905t^2 - 91.44t + 121.92 = 0$

$t = 17.19 \, \text{ sec and } \, 1.44 \, \text{ sec}$

Stone from the top of the tower:

Let t₂ = time to wait before dropping the second stone
$h = \frac{1}{2}g(t - t_2)^2$

With t = 17.19 sec
$60.96 = \frac{1}{2}(9.81)(17.19 - t_2)^2$

$t_2 = 13.67 \, \text{ sec}$

With t = 1.44 sec
$60.96 = \frac{1}{2}(32.2)(1.44 - t_2)^2$

$t_2 = -2.08 \, \text{ sec}$ (meaningless)

Use $t_2 = 13.67 \, \text{ sec}$ answer

hi, pano po nakuha yung 17.19

Use quadratic formula or use

Sir/ maam, ask ko lang po

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