Question

a ball is thrown from a height of 141 feet with an initial downward velocity of 21 ft/s. the balls height h (in feet) after t seconds is given by the following. h = 141 - 21t - 16t^2 how long after the ball is thrown does it hit the ground? round your answer(s) to the nearest hundredth. (if there is more than one answer, use the “or” button.)

Explanation:

Step1: Set height equal to 0

When the ball hits the ground, $h = 0$. So we set up the equation $0=141 - 21t-16t^{2}$.

Step2: Rearrange to standard quadratic form

We get $16t^{2}+21t - 141=0$.

Step3: Apply quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 16$, $b = 21$, and $c=-141$. First, calculate the discriminant $\Delta=b^{2}-4ac=(21)^{2}-4\times16\times(-141)=441 + 9024=9465$.

Step4: Calculate t values

$t=\frac{-21\pm\sqrt{9465}}{32}$. $\sqrt{9465}\approx97.3$. So $t=\frac{-21\pm97.3}{32}$. We have two solutions for $t$: $t_1=\frac{-21 + 97.3}{32}=\frac{76.3}{32}\approx2.38$ and $t_2=\frac{-21 - 97.3}{32}=\frac{-118.3}{32}\approx - 3.69$. Since time cannot be negative in this context, we discard the negative - valued solution.

Answer:

$t\approx2.38$ seconds