A ball is thrown into the air from the top of a cliff. The height, \( h(t) \), of the ball above the ground \( t \) seconds after it is thrown can be modeled by the following function. \( h(t)=-17 t^{2}+68 t+204 \) How many seconds after being thrown will the ball hit the ground?

To find out when the ball hits the ground, we need to determine when the height \( h(t) \) is equal to zero. We can set the equation \( -17t^2 + 68t + 204 = 0 \) and solve for \( t \) using the quadratic formula, where \( a = -17 \), \( b = 68 \), and \( c = 204 \). The formula is given by \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plugging in the values, we find the two possible solutions for \( t \). Generally, one is negative and doesn't apply here, while the positive solution represents the time when the ball reaches the ground. We can simplify the calculation; by determining the \( b^2 - 4ac \) part first: \( b^2 - 4ac = 68^2 - 4(-17)(204) = 4624 + 13824 = 18448 \). Now applying the quadratic formula gives us real-world insights into projectile motion commonly seen in sports, entertainment, and physics. This formula is commonly used in physics for solving problems related to projectile motion, allowing one to determine where an object will land after being launched. Remember, the height function can change based on variables like initial velocity, angle of launch, and environmental factors—so different scenarios could further engage your problem-solving skills and creativity in real life.