1004 Relative velocity | Rectilinear Translation

From A to C (free-fall)
$h_1 = \frac{1}{2}gt^2$

$h_1 = \frac{1}{2}(32.2)t^2$

$h_1 = 16.1t^2$
 

From B to C (upward motion)
From the formula s = v_it + ½ at²
$h_2 = v_Bt - \frac{1}{2}gt^2$

$h_2 = 40t - \frac{1}{2}(32.2)t^2$

$h_2 = 40t - 16.1t^2$
 

A to C plus B to C is equal to height of the tower
$h_1 + h_2 = h$

$16.1t^2 + (40t - 16.1t^2) = 80$

$40t = 80$

$t = 2 \, \text{ sec}$
 

$h_1 = 16.1(2^2)$

$h_1 = 64.4 \, \text{ ft}$
 

They pass each other after 2 seconds at 64.4 ft from the top of the tower.       answer
 

Velocity at C of stone from A (after 2 seconds)
$v_{C1} = gt = 32.2(2)$

$v_{C1} = 64.4 \, \text{ ft/s}$
 

Velocity at C of stone from B (after 2 seconds)
$v_{C2} = v_B - gt = = 40 - 32.2(2)$

$v_{C2} = -24.4 \, \text{ ft/s}$   →   the negative sign indicates that the stone is moving downward
 

Relative velocity:
$v_r = v_{C1} + v_{C2} = 64.4 - 24.4$

$v_r = 40 \, \text{ ft/sec}$           answer

From A to C (free-fall)
$h_1 = \frac{1}{2}gt^2$

$h_1 = \frac{1}{2}(9.81)t^2$

$h_1 = 4.905t^2$
 

From B to C (upward motion)
From the formula s = v_it + ½ at²
$h_2 = v_Bt - \frac{1}{2}gt^2$

$h_2 = 12.19t - \frac{1}{2}(9.81)t^2$

$h_2 = 12.19t - 4.905t^2$
 

A to C plus B to C is equal to height of the tower
$h_1 + h_2 = h$

$4.905t^2 + (12.19t - 4.905t^2) = 24.38$

$12.19t = 24.38$

$t = 2 \, \text{ sec}$
 

$h_1 = 4.905(2^2)$

$h_1 = 19.62 \, \text{ m}$
 

They pass each other after 2 seconds at 19.62 m from the top of the tower.       answer
 

Velocity at C of stone from A (after 2 seconds)
$v_{C1} = gt = 9.81(2)$

$v_{C1} = 19.62 \, \text{ m/s}$
 

Velocity at C of stone from B (after 2 seconds)
$v_{C2} = v_B - gt = = 12.19 - 9.81(2)$

$v_{C2} = -7.43 \, \text{ m/s}$   →   the negative sign indicates that the stone is now moving downward
 

Relative velocity:
$v_r = v_{C1} + v_{C2} = 19.62 - 7.43$

$v_r = 12.19 \, \text{ m/sec}$           answer