A ball is thrown upwards with an initial velocity of 20 m/s from a height of 50 meters. Using the equation \( h(t) = -5t^2 + 20t + 50 \), where \( h \) is the height in meters and \( t \) is the time in seconds, find the time when the ball hits the ground.

["Title: How Long Does a Ball Stay in the Air When Thrown Upwards from 50 Meters? Solving for Impact Time with a Quadratic Equation", "When a ball is thrown upward with an initial velocity and an initial height, physics meets algebra to uncover fascinating insights. In this article, we break down a classic physics scenario: a ball thrown upward from 50 meters with an initial velocity of 20 m/s, using the height equation ( h(t) = -5t^2 + 20t + 50 ). We’ll walk through the math to find exactly when the ball returns to the ground—essentially, solving for the time when ( h(t) = 0 ). Whether you're a student, teacher, or Anyone curious about kinematics, this guide demystifies the process step by step.", "---", "### The Physics Behind the Problem", "The height ( h(t) ) of the ball at time ( t ) is modeled by the quadratic equation:\n[\nh(t) = -5t^2 + 20t + 50\n]", "Why the coefficients?\n- The coefficient (-5) comes from the acceleration due to gravity (assuming downward acceleration ( g \approx 10 , \ ext{m/s}^2 ), and ( h = \frac{1}{2}gt^2 ) scaled: ( \frac{1}{2} \ imes 10t^2 = 5t^2 ), but since gravity pulls down, the equation symbols this as ( -5t^2 )),\n- The term ( +20t ) represents the initial upward velocity,\n- And ( +50 ) reflects the ball’s starting height of 50 meters.", "Our goal: find the time(s) ( t > 0 ) when the ball hits the ground, i.e., when ( h(t) = 0 ).", "---", "### Setting Up the Equation: When Does the Ball Hit the Ground?", "Set ( h(t) = 0 ):\n[\n-5t^2 + 20t + 50 = 0\n]", "Divide both sides by (-5) to simplify:\n[\nt^2 - 4t - 10 = 0\n]", "This is a standard quadratic equation in the form ( at^2 + bt + c = 0 ), where:\n- ( a = 1 )\n- ( b = -4 )\n- ( c = -10 )", "---", "### Solving the Quadratic Equation", "Use the quadratic formula:\n[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nSubstitute values:\n[\nt = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-10)}}{2(1)} = \frac{4 \pm \sqrt{16 + 40}}{2} = \frac{4 \pm \sqrt{56}}{2}\n]", "Simplify ( \sqrt{56} ):\n[\n\sqrt{56} = \sqrt{4 \ imes 14} = 2\sqrt{14}\n]", "So,\n[\nt = \frac{4 \pm 2\sqrt{14}}{2} = 2 \pm \sqrt{14}\n]", "---", "### Choosing the Valid Solution", "We have two solutions:\n[\nt = 2 + \sqrt{14} \quad \ ext{and} \quad t = 2 - \sqrt{14}\n]", "Calculate approximations:\n( \sqrt{14} \approx 3.74 )\nSo:\n- ( t \approx 2 + 3.74 = 5.74 ) seconds\n- ( t \approx 2 - 3.74 = -1.74 ) seconds (invalid, time cannot be negative)", "Only the positive value makes physical sense:\n[\n\boxed{t = 2 + \sqrt{14} \approx 5.74 \ ext{ seconds}}\n]", "---", "### Interpreting the Result", "The ball reaches the ground after approximately 5.74 seconds, starting from its initial throw. This includes both the ascent and the descent, giving a clear, accurate picture of its motion under gravity.", "---", "### Why This Equation Matters", "This kind of quadratic model is fundamental in physics and engineering. It shows how initial conditions—like initial velocity and starting height—combine with constant acceleration (gravity) to predict motion. Such equations help design safer sports fields, optimize projectile training, and educate learners in STEM.", "---", "### Final Thoughts", "Understanding when a ball hits the ground isn’t just about numbers—it’s about visualizing physics in real time. With the equation ( h(t) = -5t^2 + 20t + 50 ), we precisely determined the return-to-ground moment at approximately 5.74 seconds. Whether you’re coaching a rugby jump, analyzing a baseball pitch, or exploring kinematics, mastering these models builds deeper scientific intuition.", "---", "Keywords:\nball thrown upward, initial velocity 20 m/s, height equation ( h(t) = -5t^2 + 20t + 50 ), solving quadratic for time, physics math, root of height equation, projectile motion, ground impact time, upward throw CAD, kinematics example, motion under gravity.", "Meta Description:\nDiscover how long a ball thrown upward from 50 meters at 20 m/s stays in the air using the equation ( h(t) = -5t^2 + 20t + 50 ). We solve for the time when ( h(t) = 0 ) and explain the physics behind the quadratics. Perfect for students and physics enthusiasts."]