# 1009 Initial velocity of the second ball | Rectilinear Translation

**Problem 1009**

A ball is shot vertically into the air at a velocity of 193.2 ft per sec (58.9 m per sec). After 4 sec, another ball is shot vertically into the air. What initial velocity must the second ball have in order to meet the first ball 386.4 ft (117.8 m) from the ground?

**Solution: English System of Units**

$a = -g = -32.2 \, \text{ ft/s}^2$

$s = 386.4 \, \text{ ft}$

Thus,

$386.4 = v_it - 16.1t^2$

First ball:

$386.4 = 193.2t – 16.1t^2$

$t^2 - 12t + 24 = 0$

$t = 9.46 \, \text{ and } \, 2.5$

Use $t = 9.46 \, \text{ s}$

Second ball:

$386.4 = v_i(t - 4) - 16.1(t - 4)^2$

$386.4 = v_i(9.46 - 4) - 16.1(9.46 - 4)^2$

$v_i = 158.67 \, \text{ ft/s}$ *answer*

**Solution: International System of Units**

$a = -g = -9.81 \, \text{ m/s}^2$

$s = 117.8 \, \text{ ft}$

Thus,

$117.8 = v_it - 4.905t^2$

First ball:

$117.8 = 58.9t – 4.905t^2$

$4.905t^2 - 58.9t + 117.8 = 0$

$t = 9.47 \, \text{ and } \, 2.54$

Use $t = 9.47 \, \text{ s}$

Second ball:

$117.8 = v_i(t - 4) - 4.905(t - 4)^2$

$117.8 = v_i(9.47 - 4) - 4.905(9.47 - 4)^2$

$v_i = 48.36 \, \text{ m/s}$ *answer*