1007 Finding when and where the stones pass each other | Rectilinear Translation

$h = 1000 \, \text{ ft}$

$v_{i2} = 248 \, \text{ ft/s}$

$g = 32 \, \text{ ft/s}^2$
 

Stone dropped from captive balloon (free-falling body):
$h_1 = \frac{1}{2}gt^2 = \frac{1}{2}(32){t_1}^2$

$h_1 = 16{t_1}^2$
 

Stone thrown vertically from the ground 2 seconds later
$s = v_{i2}t_2 - \frac{1}{2}g{t_2}^2$

$h_2 = 248t_2 - \frac{1}{2}(32){t_2}^2$

$h_2 = 248(t_1 - 2) - 16(t_1 - 2)^2$
 

$h_1 + h_2 = h$

$16{t_1}^2 + [ \, 248(t_1 - 2) - 16(t_1 - 2)^2 \, ] = 1000$

$16{t_1}^2 + 248(t_1 - 2) - 16({t_1}^2 - 4t_1 + 4) = 1000$

$16{t_1}^2 + 248t_1 - 496 - 16{t_1}^2 + 64t_1 - 64 = 1000$

$16{t_1}^2 + 248t_1 - 496 - 16{t_1}^2 + 64t_1 - 64 = 1000$

$312t_1 = 1560$

$t_1 = 5 \, \text{ sec}$
 

The stones will pass each other 5 seconds after the first stone was dropped from the captive balloon.       answer
 

$h_2 = 248(5 - 2) - 16(5 - 2)^2$

$h_2 = 600 \, \text{ ft}$
 

The stones will meet at a point 600 ft above the ground.       answer

$h = 304.8 \, \text{ m}$

$v_{i2} = 75.6 \, \text{ m/s}$

$g = 9.75 \, \text{ ft/s}^2$
 

Stone dropped from captive balloon (free-falling body):
$h_1 = \frac{1}{2}gt^2 = \frac{1}{2}(9.75){t_1}^2$

$h_1 = 4.875{t_1}^2$
 

Stone thrown vertically from the ground 2 seconds later
$s = v_{i2}t_2 - \frac{1}{2}g{t_2}^2$

$h_2 = 75.6t_2 - \frac{1}{2}(9.75){t_2}^2$

$h_2 = 75.6(t_1 - 2) - 4.875(t_1 - 2)^2$
 

$h_1 + h_2 = h$

$4.875{t_1}^2 + [ \, 75.6(t_1 - 2) - 4.875(t_1 - 2)^2 \, ] = 304.8$

$4.875{t_1}^2 + 75.6(t_1 - 2) - 4.875({t_1}^2 - 4t_1 + 4) = 304.8$

$4.875{t_1}^2 + 75.6t_1 - 151.2 - 4.875{t_1}^2 + 19.5t_1 - 19.5 = 304.8$

$4.875{t_1}^2 + 75.6t_1 - 151.2 - 4.875{t_1}^2 + 19.5t_1 - 19.5 = 304.8$

$95.1t_1 = 475.5$

$t_1 = 5 \, \text{ sec}$
 

The stones will pass each other 5 seconds after the first stone was dropped from the captive balloon.       answer
 

$h_2 = 75.6(5 - 2) - 4.875(5 - 2)^2$

$h_2 = 182.925 \, \text{ m}$
 

The stones will meet at a point 182.925 m above the ground.       answer