Question

you drop a ball from a height of 98 feet. at the same time, your friend throws a ball upward. the polynomials represent the heights(in feet) of the balls after t seconds.

Answer 1

a) $h_0 -u_y t$

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:

$g=32 ft/s^2$

in the downward direction.

For the ball dropped from the initial height of $h_0 = 98 ft$, the height at time t is given by

$h(t) = h_0 -\frac{1}{2}gt^2$ (1)

The ball which is thrown upward from the ground instead is fired with an initial vertical velocity $u_y$, and its starting height is zero, so its position at time t is given by

$h'(t)=u_y t - \frac{1}{2}gt^2$ (2)

Therefore, the polynomial that represents the distance between the two balls is:

$h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t$

b)

Now we interpret this polynomial, which is:

$\Delta h(t) = h_0 -u_y t$

which represents the distance between the two balls at time t.

The interpretation of the two terms is the following:

- The constant term, $h_0$, 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, $h_0 = 98 ft$

- The coefficient of the linear term, $u_y$, is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by $u_y$ feet.