# 1010 Time to wait in dropping a stone | Rectilinear Translation

**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:

_{2}= 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**

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

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

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

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

Stone from the top of the tower:

_{2}= 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*