A projectile is fired directly upwards at 49.4 m/s. A second projectile is dropped from rest at some higher elevation at the instant the first projectile is fired and passes the first projectile 3.00 s later. From the frame of reference of the first projectile, what is the velocity of the second projectile as it passes by

Explanation:

velocity of first projectile after 3 s

v = u - gt

v = 49.4 - 9.8 x 3

= 20 m /s

Velocity of second projectile after 3 s after being dropped from rest

v = u + gt

= 0 + 9.8 x 3

= 29.4 m /s

They will be moving in opposite direction at the time of meeting , so their relative velocity

= 20 + 29.4 = 49.4 m /s

From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..