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
$h_1 = v_{i1}t - \frac{1}{2}gt^2$
$600 - 200 = 300t - \frac{1}{2}32.2t^2$
$16.1t^2 - 300t + 400 = 0$
$t = 17.19 \, \text{ sec and } \, 1.44 \, \text{ sec}$
Stone from the top of the tower:
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
$182.88 - 60.96 = 91.44t - \frac{1}{2}9.81t^2$
$4.905t^2 - 91.44t + 121.92 = 0$
With t = 17.19 sec $60.96 = \frac{1}{2}(9.81)(17.19 - t_2)^2$
With t = 1.44 sec $60.96 = \frac{1}{2}(32.2)(1.44 - t_2)^2$
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