a) [tex]h_0 -u_y t[/tex]
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:
[tex]g=32 ft/s^2[/tex]
in the downward direction.
For the ball dropped from the initial height of [tex]h_0 = 98 ft[/tex], the height at time t is given by
[tex]h(t) = h_0 -\frac{1}{2}gt^2[/tex] (1)
The ball which is thrown upward from the ground instead is fired with an initial vertical velocity [tex]u_y[/tex], and its starting height is zero, so its position at time t is given by
[tex]h'(t)=u_y t - \frac{1}{2}gt^2[/tex] (2)
Therefore, the polynomial that represents the distance between the two balls is:
[tex]h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t[/tex]
b)
Now we interpret this polynomial, which is:
[tex]\Delta h(t) = h_0 -u_y t[/tex]
which represents the distance between the two balls at time t.
The interpretation of the two terms is the following:
- The constant term, [tex]h_0[/tex], 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, [tex]h_0 = 98 ft[/tex]
- The coefficient of the linear term, [tex]u_y[/tex], is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by [tex]u_y[/tex] feet.
