A ball is thrown upward with initial velocity 20 m/s from a height of 10 m. Its height at time \(t\) is \(h(t) = -5t^2 + 20t + 10\). When does it hit the ground?

["Title: When Does a Ball Hit the Ground? Solving Its Physics Trajectory", "Understanding projectile motion is fundamental in physics and engineering, yet grasping the exact moment a ball reaches the ground often involves math and real-world application. Consider a simple yet illustrative example: a ball thrown upward with an initial velocity of 20 m/s from a height of 10 meters. Using the height equation ( h(t) = -5t^2 + 20t + 10 ), we’ll determine precisely when the ball hits the ground.", "---", "### The Equation of Motion: When Does ( h(t) = 0 )?", "The height of the ball at time ( t ) seconds is modeled by:\n[ h(t) = -5t^2 + 20t + 10 ]", "To find when the ball returns to the ground, solve for ( t ) when ( h(t) = 0 ):\n[\n-5t^2 + 20t + 10 = 0\n]", "This is a quadratic equation in standard form ( at^2 + bt + c = 0 ), with coefficients:\n- ( a = -5 )\n- ( b = 20 )\n- ( c = 10 )", "---", "### Solving the Quadratic Equation", "Use the quadratic formula:\n[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Substitute values:\n[\nt = \frac{-20 \pm \sqrt{20^2 - 4(-5)(10)}}{2(-5)} = \frac{-20 \pm \sqrt{400 + 200}}{-10} = \frac{-20 \pm \sqrt{600}}{-10}\n]", "Simplify ( \sqrt{600} ):\n[\n\sqrt{600} = \sqrt{100 \cdot 6} = 10\sqrt{6} \approx 24.49\n]", "Now compute the two possible solutions:\n[\nt = \frac{-20 + 24.49}{-10} \quad \ ext{and} \quad t = \frac{-20 - 24.49}{-10}\n]", "Calculate each:\n1. ( t = \frac{4.49}{-10} = -0.449 ) seconds (not physically meaningful, since time cannot be negative)\n2. ( t = \frac{-44.49}{-10} = 4.449 ) seconds", "---", "### When Does the Ball Hit the Ground?", "The positive solution, ( t \approx 4.45 ) seconds, is the exact time the ball lands on the ground.", "---", "### Physical Interpretation", "- The ball starts at 10 m, rises under gravity, and eventually falls.\n- The quadratic shape reflects acceleration due to gravity (negative quadratic term).\n- The vertex of the parabola marks the peak height, after which the ball descends.\n- Only the positive time matters—mathematically, the ball returns just once after leaving the ground.", "---", "### Final Answer", "The ball hits the ground after approximately 4.45 seconds.", "Understanding this calculation helps predict motion in sports, engineering, and everyday physics problems. Use ( h(t) = -5t^2 + 20t + 10 ) and the quadratic formula to solve similar problems anywhere a projectile follows a parabolic trajectory under constant gravity.", "---", "Keywords: ball thrown upward, height equation h(t), projectile motion, quadratic formula physics, when does ball hit ground, physics problem solution, earth gravitational motion", "Meta Description: Use the equation ( h(t) = -5t^2 + 20t + 10 ) to find when a 20 m/s upward throw from 10 m hits the ground. Solution explained with quadratics and real-world context."]