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

["Title: When Does a Ball Throw Specific Distance Hit the Ground? | Solving the Physics Problem", "Meta Description: Want to know when a ball thrown upward at 20 m/s from 2 meters will hit the ground? Learn how to solve the equation ( h(t) = -5t^2 + 20t + 2 ) and find the exact time using simple physics and algebra.", "---", "### Problem: A Ball Thrown Upward Hits the Ground\nWhen thrown upward with an initial velocity of 20 m/s from a height of 2 meters, the height of the ball at time ( t ) seconds is modeled by the quadratic equation:", "[\nh(t) = -5t^2 + 20t + 2\n]", "But an important question arises: At what time ( t ) does the ball strike the ground?", "### Understanding the Equation\nThis equation represents motion under gravity, where the maximum height and descent follow a parabolic path. The coefficient ( -5 ) comes from the acceleration due to gravity (( -9.8 , \ ext{m/s}^2 )), scaled to approximate ( -½ \ imes 9.8 ) for easier quadratic calculations in meters and seconds. The constant ( +2 ) accounts for the starting height.", "We’re solving for when the height ( h(t) ) becomes zero — the moment the ball touches the ground.", "### Setting Up the Equation\nSet ( h(t) = 0 ):", "[\n-5t^2 + 20t + 2 = 0\n]", "This is a standard quadratic equation of the form ( at^2 + bt + c = 0 ), where:\n- ( a = -5 )\n- ( b = 20 )\n- ( c = 2 )", "### Solving the Quadratic Equation\nUse the quadratic formula:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Plug in the values:", "[\nt = \frac{-20 \pm \sqrt{20^2 - 4(-5)(2)}}{2(-5)} = \frac{-20 \pm \sqrt{400 + 40}}{-10} = \frac{-20 \pm \sqrt{440}}{-10}\n]", "Simplify ( \sqrt{440} ):", "[\n\sqrt{440} = \sqrt{4 \ imes 110} = 2\sqrt{110} \approx 2 \ imes 10.488 = 20.976\n]", "Now compute the two possible solutions:", "[\nt = \frac{-20 + 20.976}{-10} = \frac{0.976}{-10} = -0.0976 \quad \ ext{(not physically meaningful, time cannot be negative)}\n]", "[\nt = \frac{-20 - 20.976}{-10} = \frac{-40.976}{-10} = 4.0976 \approx 4.10 \ ext{ seconds}\n]", "### Interpretation\nThe ball hits the ground at approximately 4.10 seconds after being thrown. The negative root is dismissed because time cannot be negative in this context.", "### Why This Matters\nThis calculation demonstrates how quadratic equations model real-world phenomena such as projectile motion. Accurately predicting when an object returns to the ground helps in sports, engineering, physics education, and safety planning.", "---", "### Final Answer:\nThe ball hits the ground at approximately 4.10 seconds.", "Use this method to analyze any vertical throw: plug height equation to zero and apply the quadratic formula for precise timing. Whether for classroom learning or practical application, mastering this approach builds strong problem-solving skills in physics and applied math.", "---", "Keywords:\nprojectile motion, height equation, quadratic formula, time to ground, ball thrown upward, physics problem solving, kinematics, ground impact time, motion in two dimensions", "Related Readings:\n- How to solve ( h(t) = -5t^2 + 20t + 2 = 0 )\n- Projectile motion example problems\n- Physics of vertical throws and quadratic equations", "---", "Optimized for search engines with clear headings, technical background, step-by-step solving, and practical relevance — ideal for students and educators studying kinematics and motion dynamics."]