a) ![h_0 -u_y t](https://tex.z-dn.net/?f=h_0%20-u_y%20t)
b) See interpretation below
Step-by-step explanation:
a)
The motion of both balls is a free-fall motion: it means that the ball is acted upon the force of gravity only.
Therefore, this means that the motion of the ball is a uniformly accelerated motion, with constant acceleration equal to the acceleration of gravity:
![g=32 ft/s^2](https://tex.z-dn.net/?f=g%3D32%20ft%2Fs%5E2)
in the downward direction.
For the ball dropped from the initial height of
, the height at time t is given by
(1)
The ball which is thrown upward from the ground instead is fired with an initial vertical velocity
, and its starting height is zero, so its position at time t is given by
(2)
Therefore, the polynomial that represents the distance between the two balls is:
![h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t](https://tex.z-dn.net/?f=h%28t%29-h%27%28t%29%3Dh_0%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%20-%20%28u_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%29%20%3D%20h_0%20-u_y%20t)
b)
Now we interpret this polynomial, which is:
![\Delta h(t) = h_0 -u_y t](https://tex.z-dn.net/?f=%5CDelta%20h%28t%29%20%3D%20h_0%20-u_y%20t)
which represents the distance between the two balls at time t.
The interpretation of the two terms is the following:
- The constant term,
, is the initial distance between the two balls, at time t=0 (in fact, the first ball is still at the top of the building, while the second ball is on the ground). For this problem, ![h_0 = 98 ft](https://tex.z-dn.net/?f=h_0%20%3D%2098%20ft)
- The coefficient of the linear term,
, is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by
feet.