Acceleration, velocity, distance problem

Question

Two balls are thrown upward from the edge of the cliff 432 feet above the ground. The first is thrown with a speed of 48 ft/s and the other is thrown a second later with a speed of 24 ft/s. Do the balls ever pass each other?

Answer 1

ho = 432 = initial height for both, vo= 48 ft/sec = 1st ball's initial velocity

2nd vo=24 ft/sec = initial velocity for 2nd ball

-16t^2 reflects the negative acceleration due to gravity

general equation is

h(t) = (a/2)t^2 +vot +ho were a = -32 ft/sec^2 on earth at sea level approximately

1st ball: h(t) =-16t^2 +48t +432. 2nd ball:h(t) -16t^2 +24t +432 if they left at the same time, but 2nd ball leaves 1 second later, which is a rightward shift of 1, so rewrite 2nd parabola as -16(t-1)^2 +48(t-1) = 432

or simplified to h(t) -16t^2 + 56t +392

set the two parabolic equations equal and solve for t

-16^t^2 +56t + 392 = -16t^2 +48t + 432

8t = 40

t= 5 seconds for the two balls to be at the same height

h(5) = -16(25)+56(5) +392

= -640 + 280 +392

= 32 feet high for 2nd ball in 5 seconds

h(5) = -16(25)+48(5) +432

=-640 +240 +432

= 32 feet high for 1st ball in 5 seconds

the two balls pass each other at 32 feet above ground or 400 feet below the cliff where they started, in 5 seconds. the 1st ball is in the air 5 seconds, the 2nd is in the air for (5-1)= 4 seconds