# 1004 Relative velocity | Rectilinear Translation

**Problem 1004**

A ball is dropped from the top of a tower 80 ft (24.38 m) high at the same instant that a second ball is thrown upward from the ground with an initial velocity of 40 ft/sec (12.19 m/s). When and where do they pass, and with what relative velocity?

**Solution in English System**

## Click here to expand or collapse this section

v

_{A}= 0

v

_{B}= 40 ft/s

g = 32.2 ft/s

^{2}

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_{i}t + ½ at^{2}

$h_2 = v_Bt - \frac{1}{2}gt^2$

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

$h_2 = 40t - 16.1t^2$

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*

**Solution in SI Units**

## Click here to expand or collapse this section

v

_{A}= 0

v

_{B}= 12.19 m/s

g = 9.81 m/s

^{2}

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_{i}t + ½ at^{2}

$h_2 = v_Bt - \frac{1}{2}gt^2$

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

$h_2 = 12.19t - 4.905t^2$

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*

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