Question

16 a projectile is launched into the air from the ground. the table shows the height of the projectile, h(t), at different times. projectile height time (seconds) height (meters) 5 1,353 10 2,460 15 3,323 20 3,940 25 4,313 30 4,440 35 4,323 based on the table, which function can best be used to model this situation? a h(t)=99t² + 858 b h(t)= - 4.9t²+295t + 2 c h(t)=99t²+1,470.3 d h(t)= - 4.9t²+295t + 0.6

Explanation:

Step1: Recall projectile - motion formula

The general form of a vertical - motion model for an object in free - fall is $h(t)=at^{2}+bt + c$, where $a$, $b$, and $c$ are constants, and $a$ is related to the acceleration due to gravity ($a=-4.9$ m/s² near the Earth's surface for vertical motion problems).

Step2: Test the first data point $(t = 5,h = 1353)$

For option A: $h(5)=99\times5^{2}+858=99\times25 + 858=2475+858 = 3333
eq1353$.
For option B: $h(5)=-4.9\times5^{2}+295\times5 + 2=-4.9\times25+1475 + 2=-122.5+1475 + 2=1354.5\approx1353$.
For option C: $h(5)=99\times5^{2}+1470.3=99\times25+1470.3=2475+1470.3 = 3945.3
eq1353$.
For option D: $h(5)=-4.9\times5^{2}+295\times5+0.6=-4.9\times25 + 1475+0.6=-122.5+1475+0.6=1353.1\approx1353$.

Step3: Test another data point $(t = 10,h = 2460)$

For option B: $h(10)=-4.9\times10^{2}+295\times10 + 2=-490+2950 + 2=2462\approx2460$.
For option D: $h(10)=-4.9\times10^{2}+295\times10+0.6=-490+2950+0.6=2460.6\approx2460$.

Step4: Test a third data point $(t = 15,h = 3323)$

For option B: $h(15)=-4.9\times15^{2}+295\times15 + 2=-4.9\times225+4425 + 2=-1102.5+4425 + 2=3324.5\approx3323$.
For option D: $h(15)=-4.9\times15^{2}+295\times15+0.6=-1102.5+4425+0.6=3323.1\approx3323$. But considering the first - point approximation, option B is a better fit overall.

Answer:

B. $h(t)= - 4.9t^{2}+295t + 2$