Question

5 multiple choice 16.5 points
a firework is launched into the air from ground level with an initial velocity of 128 ft/s. if acceleration due to gravity is -16 ft/s², what is the maximum height reached by the
firework?
h(t) = -16t² + 128t
448 ft
256 ft
512 ft
1,024 ft
6 multiple choice 16.5 points
a softball pitcher throws a softball to a catcher behind home plate. the softball is thrown at a velocity of 50 feet per second. if the ball can be modeled by the following
equation, when is the ball moving through the air?
v(t) = -16t² + 50t
0<t<4.25
-3.125<t<0
0<t<3.125
-2<t<3.125

Explanation:

Question 5

Step1: Find the time at maximum height

For a quadratic function \(h(t)=at^{2}+bt + c\) (here \(a=-16\), \(b = 128\), \(c = 0\)), the time \(t\) at the vertex (maximum for \(a<0\)) is given by \(t=-\frac{b}{2a}\).

$$t=-\frac{128}{2\times(-16)}=\frac{- 128}{-32}=4$$

Step2: Calculate the maximum height

Substitute \(t = 4\) into \(h(t)=-16t^{2}+128t\).

$$h(4)=-16\times4^{2}+128\times4=-16\times16 + 512=-256+512 = 256$$

Step1: Find when \(h(t)=0\)

Set \(h(t)=-16t^{2}+50t = 0\). Factor out \(t\): \(t(-16t + 50)=0\).
We get \(t = 0\) (initial time) and \(-16t+50=0\). Solving \(-16t+50 = 0\) for \(t\):

$$16t=50\Rightarrow t=\frac{50}{16}=3.125$$

The ball is in the air when \(h(t)>0\), so \(0 < t<3.125\)

Answer:

256 ft

Question 6