Question

a baseball is thrown into the air from a height of 5 feet. the ball reaches a maximum height of 43.5 feet and spends a total of 3.2 seconds in the air. which equation models the height of the baseball? assume that acceleration due to gravity is $-16$ ft/s$^2$.

$\bigcirc\\ h(t) = 16t^2 + 49.64t + 5$
$\bigcirc\\ h(t) = -16t^2 + 5t + 49.64$
$\bigcirc\\ h(t) = -16t^2 + 49.64t + 5$
$\bigcirc\\ h(t) = 16t^2 + 5t + 49.64$

Explanation:

Step1: Recall the projectile motion formula

The general formula for the height of a projectile is \( h(t) = \frac{1}{2}at^2 + v_0t + h_0 \), where \( a \) is the acceleration, \( v_0 \) is the initial velocity, and \( h_0 \) is the initial height. Given \( a=- 16\space ft/s^{2}\), so \( \frac{1}{2}a=- 8\)? Wait, no, wait, the standard formula for vertical motion under gravity is \( h(t)=-16t^{2}+v_0t + h_0\) (since \( a = - 32\space ft/s^{2}\) for gravity, wait, wait, the problem says acceleration due to gravity is \(- 16\space ft/s^{2}\)? Wait, maybe the problem is using a simplified model where \( a=-16\), so the formula is \( h(t)=-16t^{2}+v_0t + h_0\). The initial height \( h_0 = 5\) feet (thrown from 5 feet), so the constant term should be 5. So we can eliminate options where the constant term is not 5. So options with constant term 49.64 (second and fourth options) are out. Now we have first and third options.

Step2: Analyze the sign of the quadratic coefficient

Since the ball is thrown upward and then comes down, the parabola should open downward, so the coefficient of \( t^{2}\) should be negative. So the first option has \( 16t^{2}\) (positive, opening upward) which is wrong. So we are left with the third option: \( h(t)=-16t^{2}+49.64t + 5\). Let's verify the maximum height. The time at maximum height for \( h(t)=at^{2}+bt + c\) is \( t=-\frac{b}{2a}\). Here \( a=-16\), \( b = 49.64\), so \( t=-\frac{49.64}{2\times(-16)}=\frac{49.64}{32}\approx1.55125\) seconds. Then \( h(1.55125)=-16\times(1.55125)^{2}+49.64\times1.55125 + 5\). Calculate \( (1.55125)^{2}\approx2.4064\), so \(-16\times2.4064\approx - 38.5024\). \( 49.64\times1.55125\approx49.64\times1.5 + 49.64\times0.05125\approx74.46+2.543\approx77.003\). Then \( h(1.55125)\approx-38.5024 + 77.003+5\approx43.5006\), which matches the maximum height of 43.5 feet. Also, the total time in air is 3.2 seconds. Let's check when \( h(t) = 0\) (ground level). \( -16t^{2}+49.64t + 5=0\). Using quadratic formula \( t=\frac{-49.64\pm\sqrt{49.64^{2}-4\times(-16)\times5}}{2\times(-16)}\). Calculate discriminant \( D = 49.64^{2}+320\approx2464.1296 + 320=2784.1296\). \(\sqrt{D}\approx52.76\). Then \( t=\frac{-49.64\pm52.76}{-32}\). We take the positive root: \( t=\frac{-49.64 + 52.76}{-32}=\frac{3.12}{-32}\) (no, wait, wait, \( a=-16\), so denominator is \( 2a=-32\). Wait, \( t=\frac{-b\pm\sqrt{D}}{2a}=\frac{-49.64\pm52.76}{-32}\). So two roots: \( t_1=\frac{-49.64 + 52.76}{-32}=\frac{3.12}{-32}\approx - 0.0975\) (discard, time can't be negative), \( t_2=\frac{-49.64-52.76}{-32}=\frac{-102.4}{-32}=3.2\) seconds. Which matches the total time in air (3.2 seconds). So the correct equation is \( h(t)=-16t^{2}+49.64t + 5\).

Answer:

\( h(t) = -16t^2 + 49.64t + 5 \) (the third option)