1009 Initial velocity of the second ball | Rectilinear Translation

$s = v_it + \frac{1}{2}at^2$
 

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

$s = 386.4 \, \text{ ft}$
 

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

$s = v_it + \frac{1}{2}at^2$
 

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

$s = 117.8 \, \text{ ft}$
 

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