$s = v_it + \frac{1}{2}at^2$

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

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

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*

## Comments

## hi.po good day.paano po

hi.po good day.paano po naging 9.47s sa time? thank you in advance po sa sasagot

## 9.47 was found by quadratic

9.47 was found by quadratic formula, I believe you get that, I think what you mean is, why use 9.47 and not 2.54 sec for

t. Using 2.54 sec fortis invalid because the second ball was shot 4 seconds after the first ball.## thank you po. na checked ko

thank you po. na checked ko na po using shft.solve