Question

cp algebra 2
projectile motion problems
show all work to receive full credit. round answers to the nearest hundredth.

1. we threw a turkey upward from the top of a building. the height of the object, h (measured in feet) t seconds after

we threw it is h(t) = -16t² + 160t + 120.
a. when is the turkey 200 feet above the ground?
b. when will the turkey land on the ground?
c. when is the turkey at its maximum height? what is the maximum height?

1. an athlete throws a discus from an initial height of 6 feet and an initial vertical velocity of 46 feet per second.

a. after how many seconds does the discus hit the ground?
b. can the discus reach a height of 45 feet?
c. what is the maximum height?

1. a rocketry club is launching model rockets. the launching pad is 30 feet above the ground. your model rocket has an

initial vertical velocity of 105 feet per second. your friend’s model rocket has an initial vertical velocity of 100 feet per
second.
a. when does your rocket reach a height of 200 feet?
b. when does your friend’s rocket reach the same height?
b. which rocket was in the air longer?

Explanation:

Step 1: Solve part 1a

We set \( h(t)=200 \), so the equation is \( -16t^{2}+160t + 120=200 \).
Subtract 200 from both sides: \( -16t^{2}+160t - 80 = 0 \).
Divide all terms by -16: \( t^{2}-10t + 5=0 \).
Using the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 1 \), \( b=- 10 \), \( c = 5 \).
\( t=\frac{10\pm\sqrt{100 - 20}}{2}=\frac{10\pm\sqrt{80}}{2}=\frac{10\pm4\sqrt{5}}{2}=5\pm2\sqrt{5} \).
\( 5 + 2\sqrt{5}\approx5 + 4.47 = 9.47 \), \( 5-2\sqrt{5}\approx5 - 4.47 = 0.53 \).

Step 2: Solve part 1b

We set \( h(t) = 0 \), so \( -16t^{2}+160t + 120=0 \).
Divide by -8: \( 2t^{2}-20t - 15 = 0 \).
Using quadratic formula \( t=\frac{20\pm\sqrt{400+120}}{4}=\frac{20\pm\sqrt{520}}{4}=\frac{20\pm2\sqrt{130}}{4}=\frac{10\pm\sqrt{130}}{2} \).
\( \sqrt{130}\approx11.40 \), so \( t=\frac{10 + 11.40}{2}\approx10.70 \) (we discard the negative solution).

Step 3: Solve part 1c

For a quadratic \( ax^{2}+bx + c \), the time of maximum height is \( t=-\frac{b}{2a} \). Here \( a=-16 \), \( b = 160 \), so \( t=-\frac{160}{2\times(-16)} = 5 \) seconds.
Maximum height \( h(5)=-16(25)+160(5)+120=-400 + 800+120 = 520 \) feet.

Step 4: Solve part 2a

The height function for the discus is \( h(t)=-16t^{2}+46t + 6 \). Set \( h(t)=0 \).
Using quadratic formula \( t=\frac{-46\pm\sqrt{46^{2}-4\times(-16)\times6}}{2\times(-16)}=\frac{-46\pm\sqrt{2116 + 384}}{-32}=\frac{-46\pm\sqrt{2500}}{-32}=\frac{-46\pm50}{-32} \).
We take the positive solution: \( t=\frac{-46 + 50}{-32}=\frac{4}{-32}=-0.125 \) (discard), \( t=\frac{-46-50}{-32}=\frac{-96}{-32}=3 \) seconds.

Step 5: Solve part 2b

Time of maximum height \( t=-\frac{46}{2\times(-16)}=\frac{46}{32}=\frac{23}{16}\approx1.44 \) seconds.
Maximum height \( h(\frac{23}{16})=-16(\frac{23}{16})^{2}+46(\frac{23}{16})+6=- \frac{529}{16}+\frac{1058}{16}+6=\frac{529}{16}+6=\frac{529 + 96}{16}=\frac{625}{16}=39.0625\lt45 \), so no.

Step 6: Solve part 2c

From part 2b, maximum height is \( \frac{625}{16}=39.06 \) feet (or 39.06 feet).

Step 7: Solve part 3a

Your rocket's height function: \( h(t)=-16t^{2}+105t + 30 \). Set \( h(t)=200 \).
\( -16t^{2}+105t + 30 - 200=0\Rightarrow -16t^{2}+105t - 170 = 0\Rightarrow16t^{2}-105t + 170 = 0 \).
Quadratic formula: \( t=\frac{105\pm\sqrt{105^{2}-4\times16\times170}}{2\times16}=\frac{105\pm\sqrt{11025 - 10880}}{32}=\frac{105\pm\sqrt{145}}{32} \).
\( \sqrt{145}\approx12.04 \), so \( t=\frac{105 + 12.04}{32}\approx3.66 \), \( t=\frac{105 - 12.04}{32}\approx2.89 \).

Step 8: Solve part 3b

Friend's rocket: \( h(t)=-16t^{2}+100t + 30 \). Set \( h(t)=200 \).
\( -16t^{2}+100t + 30 - 200=0\Rightarrow -16t^{2}+100t - 170 = 0\Rightarrow8t^{2}-50t + 85 = 0 \).
Discriminant \( \Delta=2500-4\times8\times85=2500 - 2720=-220\lt0 \)? Wait, no, calculation error. Wait, \( -16t^{2}+100t - 170 = 0\Rightarrow16t^{2}-100t + 170 = 0\Rightarrow8t^{2}-50t + 85 = 0 \). Discriminant \( \Delta = 2500-4\times8\times85=2500 - 2720=-220 \), which is wrong. Wait, original equation: \( h(t)=-16t^{2}+100t + 30 = 200\Rightarrow -16t^{2}+100t - 170 = 0\Rightarrow16t^{2}-100t + 170 = 0 \). Divide by 2: \( 8t^{2}-50t + 85 = 0 \). Discriminant \( \Delta=(-50)^{2}-4\times8\times85=2500 - 2720=-220 \), which is negative. Wait, that means my friend's rocket never reaches 200 feet? Wait, maybe I made a mistake. Wait, initial velocity of friend is 100, mine is 105. My rocket's height function: \( h(t)=-16t^{2}+105t + 30 \). Let's check the maximum height of my rocket: \( t = \frac{105}{32}\approx3.28 \), \( h(\frac{105}{32})=-16(\frac{105}{32})^{2}+105\times\frac{105}{32}+30 \). Calculate: \( -16\times\frac{11025}{1024}+\frac{11025}{32}+30=-\frac{11025}{64}+\frac{22050}{64}+\frac{1920}{64}=\frac{-11025 + 22050+1920}{64}=\frac{12945}{64}\approx202.27 \), so it can reach 200. Friend's rocket: \( t=\frac{100}{32}=\frac{25}{8}=3.125 \), \( h(\frac{25}{8})=-16(\frac{625}{64})+100\times\frac{25}{8}+30=-\frac{625}{4}+\frac{2500}{8}+30=-\frac{1250}{8}+\frac{2500}{8}+\frac{240}{8}=\frac{1490}{8}=186.25\lt200 \). So friend's rocket never reaches 200 feet.

Step 9: Solve part 3b (the other part, which rocket is in air longer)

Your rocket: \( h(t)=-16t^{2}+105t + 30 = 0 \). Quadratic formula: \( t=\frac{-105\pm\sqrt{105^{2}-4\times(-16)\times30}}{2\times(-16)}=\frac{-105\pm\sqrt{11025 + 1920}}{-32}=\frac{-105\pm\sqrt{12945}}{-32} \). \( \sqrt{12945}\approx113.78 \), so \( t=\frac{-105 - 113.78}{-32}\approx\frac{-218.78}{-32}\approx6.84 \).
Friend's rocket: \( h(t)=-16t^{2}+100t + 30 = 0 \). Quadratic formula: \( t=\frac{-100\pm\sqrt{10000+1920}}{-32}=\frac{-100\pm\sqrt{11920}}{-32}=\frac{-100\pm109.18}{-32} \). Positive solution: \( t=\frac{-100 - 109.18}{-32}\approx\frac{-209.18}{-32}\approx6.54 \). So my rocket is in air longer.

Answer:

1a. \( t\approx0.53 \) seconds and \( t\approx9.47 \) seconds
1b. \( t\approx10.70 \) seconds
1c. At \( t = 5 \) seconds, maximum height is 520 feet
2a. \( t = 3 \) seconds
2b. No
2c. 39.06 feet
3a. \( t\approx2.89 \) seconds and \( t\approx3.66 \) seconds
3b. Your rocket (my rocket) is in the air longer (friend's rocket never reaches 200 feet)
3b (the other part): Your rocket is in the air longer (approx 6.84 seconds vs friend's approx 6.54 seconds)