To solve the given problem, we need to determine when the ball hits the ground. This happens when the height $h$ is equal to 0. The equation for the ball’s height is:

$h(t) = 197 - 21t - 16t^2$

We set $h(t) = 0$ to find the time $t$ at which the ball hits the ground:

$0 = 197 - 21t - 16t^2$

Rearrange the equation into standard quadratic form:

$-16t^2 - 21t + 197 = 0$

Multiply through by $-1$ to make the leading coefficient positive:

$16t^2 + 21t - 197 = 0$

Solve using the quadratic formula:

The quadratic formula is:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 16$, $b = 21$, and $c = -197$. Substitute these values into the formula:

$t = \frac{-21 \pm \sqrt{21^2 - 4(16)(-197)}}{2(16)}$

1. Compute the discriminant:

$\Delta = b^2 - 4ac = 21^2 - 4(16)(-197) = 441 + 12608 = 13049$

2. Take the square root of the discriminant:

$\sqrt{13049} \approx 114.2$

3. Compute the two solutions for $t$:

$t = \frac{-21 \pm 114.2}{32}$

First solution:

$t_1 = \frac{-21 + 114.2}{32} = \frac{93.2}{32} \approx 2.91$

Second solution:

$t_2 = \frac{-21 - 114.2}{32} = \frac{-135.2}{32} \approx -4.23$

Since time $t$ cannot be negative, the only valid solution is:

$t \approx 2.91 \, \text{seconds}$

Final Answer:

The ball hits the ground approximately 2.91 seconds after it is thrown.

Would you like a detailed explanation of any step or clarification?

Related Questions:

1. How does the quadratic formula work in general for solving equations?
2. Why do we discard the negative solution for time in this context?
3. What is the significance of the discriminant in the quadratic equation?
4. How does gravity influence the $t^2$-term in the height equation?
5. Can we solve this problem using a different method, such as factoring?

Tip:

Always verify the units in a physics-related problem to ensure accuracy in real-world applications!