A ball is thrown upward with an initial velocity of 20 m/s from a height of 5 meters. Its height after \( t \) seconds is modeled by \( h(t) = -5t^2 + 20t + 5 \). When does the ball hit the ground?

["Title: How High Does a Ball Go? Solving for Time When It Hits the Ground Using Physics and Quadratic Equations", "Meta Description:\nLearn how to calculate the time a thrown ball hits the ground using the height equation ( h(t) = -5t^2 + 20t + 5 ). Solve for ( t ) when ( h(t) = 0 ) to find the impact moment.", "---", "### When a ball is thrown upward with an initial velocity of 20 m/s from a height of 5 meters, its height above the ground is modeled by the quadratic equation:\n[ h(t) = -5t^2 + 20t + 5 ]\nUnderstanding when this ball hits the ground means determining the time ( t ) when ( h(t) = 0 ). In this article, we’ll walk through solving the quadratic equation, interpreting the physical meaning, and explaining how math helps predict real-world motion.", "---", "### The Physics Behind the Equation", "When an object is thrown upward, gravity pulls it down while its initial upward velocity slows it. The height function incorporates:\n- The negative coefficient on ( t^2 ) accounts for gravitational acceleration ((-5 , \ ext{m/s}^2)).\n- The ( +20t ) term reflects the initial upward momentum.\n- The constant ( +5 ) m represents the starting height.", "To find when the ball returns to ground level, set ( h(t) = 0 ):\n[\n-5t^2 + 20t + 5 = 0\n]", "---", "### Solving the Quadratic Equation", "We solve ( -5t^2 + 20t + 5 = 0 ). The quadratic formula is:\n[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nHere, ( a = -5 ), ( b = 20 ), ( c = 5 ). Plug in the values:\n[\nt = \frac{-20 \pm \sqrt{(20)^2 - 4(-5)(5)}}{2(-5)} = \frac{-20 \pm \sqrt{400 + 100}}{-10} = \frac{-20 \pm \sqrt{500}}{-10}\n]", "Simplify ( \sqrt{500} = \sqrt{100 \cdot 5} = 10\sqrt{5} \approx 22.36 ):\n[\nt = \frac{-20 \pm 22.36}{-10}\n]", "Calculate both roots:\n1. ( t = \frac{-20 + 22.36}{-10} = \frac{2.36}{-10} = -0.236 ) seconds (invalid, time can’t be negative)\n2. ( t = \frac{-20 - 22.36}{-10} = \frac{-42.36}{-10} = 4.236 ) seconds", "Thus, the ball hits the ground approximately 4.24 seconds after being thrown.", "---", "### Real-World Interpretation", "The positive root ( t \approx 4.24 ) s represents the full duration the ball is airborne. Between takeoff (( t = 0 )) and impact, it rises to a peak height before descending.", "This calculation is vital for sports, engineering, and physics simulations, helping predict motion accurately and safely.", "---", "### Summary", "Given the height equation\n[ h(t) = -5t^2 + 20t + 5 ]\nthe ball hits the ground at\n[\nt = \frac{-20 + \sqrt{500}}{-10} \approx 4.24 \ ext{ seconds}\n]\nUse the positive root and interpret it as the full flight time from release to impact.", "---", "Keywords: ball thrown upward, height equation, -5t² + 20t + 5, how long until ball hits ground, quadratic equation physics, projectile motion, solve for time, ground impact time, height function, algebraic method, motion modeling", "---", "Optimize this article with internal links to related materials like "projectile motion formulas" or "real-world physics applications," and consider adding a visual graph of the height curve crossing the ( t )-axis at impact time to improve engagement and SEO."]