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

 

Solution: International System of Units

 

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

In reply to by oggiessenpai

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.

Why use 9.47s?

In reply to by Angel54212689 (not verified)

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 - 4 ang t2 and not t2 = t1 + 4? Dito po ako palaging nagkakamali. Please enlighten me po. Thank you po.

In reply to by Gab123456789 (not verified)

consider niyo po as t yung 9.46 then yun yung time ng ball 1. then t1=4s, and in order to solve for the initial velocity of ball 2, kailangan mo yung 22 which is ma-eestablish sa relationship na t=t1+t2. that means to say, t2=t-t1=t-4