$s = v_it + \frac{1}{2}at^2$
$a = -g = -9.81 \, \text{ m/s}^2$
$s = 117.8 \, \text{ ft}$
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 for t is 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
Why use 9.47s?
Why use 9.47s?
Using 2.54 sec for t is
Using 2.54 sec for t is invalid because the second ball was shot 4 seconds after the first ball.
Hello po! Paano po naging t1
Hello po! Paano po naging t1 - 4 ang t2 and not t2 = t1 + 4? Dito po ako palaging nagkakamali. Please enlighten me po. Thank you po.