Question

if a ball is thrown into the air at 64 feet per second from the top of a 119-foot-tall building, its height can be modeled by the function $s = 119 + 64t - 16t^2$, where s is in feet and t is in seconds. complete parts a through c below

a. graph this function for t representing 0 to 8 seconds and s representing 0 to 200 feet. choose the correct graph below
a.
b.
c.
d.

b. find the height of the ball 1 second after it is thrown and 3 seconds after it is thrown. how can these values be equal?
the height of the ball 1 second after it is thrown is $square$ ft
the height of the ball 3 seconds after it is thrown is $square$ ft
how can these values be equal?
a. these two values are equal because the ball was rising to a maximum height at the first instance and then after reaching the maximum height, the ball was falling at the second instance. in the first instance, 1 second after throwing the ball in an upward direction, it will reach the height 167 ft and in the

Explanation:

Step1: Identify correct graph (part a)

The function $S(t) = 119 + 64t - 16t^2$ is a downward-opening parabola (negative $t^2$ coefficient) with initial height $S(0)=119$ ft. Only option A matches: starts above 0, peaks, then falls to 0.

Step2: Calculate height at t=1 (part b)

Substitute $t=1$ into $S(t)$:
$S(1) = 119 + 64(1) - 16(1)^2 = 119 + 64 - 16 = 167$

Step3: Calculate height at t=3 (part b)

Substitute $t=3$ into $S(t)$:
$S(3) = 119 + 64(3) - 16(3)^2 = 119 + 192 - 144 = 167$

Step4: Explain equal heights (part b)

The parabola is symmetric about its vertex. The vertex time is $t=-\frac{b}{2a} = -\frac{64}{2(-16)} = 2$ seconds. $t=1$ and $t=3$ are equally spaced around $t=2$, so heights are equal: the ball rises to the vertex at $t=2$, so at $t=1$ it is rising, and at $t=3$ it is falling to the same height.

Answer:

a. A
b.
The height of the ball 1 second after it is thrown is 167 ft
The height of the ball 3 seconds after it is thrown is 167 ft
How can these values be equal?
A. These two values are equal because the ball was rising to a maximum height at the first instance and then after reaching the maximum height, the ball was falling at the second instance. In the first instance, 1 second after throwing the ball in an upward direction, it will reach the height 167 ft and in the second instance, 3 seconds after the ball is thrown, it will come back to the height 167 ft.