A ball is thrown upwards with an initial velocity of 20 m/s from a height of 50 meters. How long will it take to hit the ground? (Use \( g = 9.8 \) m/s\(^2\))

["Title: How Long Does a Ball Take to Hit the Ground When Thrown Upward? (50m Height, 20 m/s Initial Velocity)", "---", "Ever wondered how long a ball thrown upward at 20 m/s from a 50-meter height stays in the air before landing? This classic physics problem combines kinematics and gravity to deliver a clear understanding of motion under constant acceleration. Here, we explain how to calculate the total time from throw until impact using the equation of motion — no complicated math, just physics in action.", "---", "### Scenario Overview", "A ball is thrown vertically upward with:", "- Initial velocity (( u )) = 20 m/s\n- Starting height (( s_0 )) = 50 meters\n- Acceleration due to gravity (( g )) = 9.8 m/s² (acting downward)", "The goal is to find how long it takes for the ball to hit the ground.", "---", "### Using the Kinematic Equation", "We use the equation of motion for vertical displacement under constant acceleration:", "[\ns = s_0 + ut + \frac{1}{2} a t^2\n]", "Where:\n- ( s ) = final position (ground level) = 0 meters\n- ( s_0 = 50 ) m (initial height)\n- ( u = 20 ) m/s (upward)\n- ( a = -g = -9.8 ) m/s² (acceleration due to gravity opposing the motion)", "Substitute values:", "[\n0 = 50 + 20t - \frac{1}{2}(9.8)t^2\n]", "Simplify:", "[\n0 = 50 + 20t - 4.9t^2\n]", "Rearranged:", "[\n4.9t^2 - 20t - 50 = 0\n]", "This is a quadratic equation:\n[\nat^2 + bt + c = 0\n]\nwith\n( a = 4.9 ), ( b = -20 ), ( c = -50 )", "---", "### Solving the Quadratic Equation", "Use 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(4.9)(-50)}}{2(4.9)}\n]\n[\nt = \frac{20 \pm \sqrt{400 + 980}}{9.8}\n]\n[\nt = \frac{20 \pm \sqrt{1380}}{9.8}\n]\n[\n\sqrt{1380} \approx 37.15\n]", "So:", "[\nt = \frac{20 + 37.15}{9.8} \quad \ ext{(we discard the negative root as time can't be negative)}\n]\n[\nt \approx \frac{57.15}{9.8} \approx 5.83 \ ext{ seconds}\n]", "---", "### Final Answer", "The ball will take approximately 5.83 seconds to hit the ground.", "---", "### Summary", "Solving projectile problems step-by-step helps clarify how gravity affects upward motion. By modeling vertical motion using the equations of motion and solving the resulting quadratic, we find the precise time until impact. Whether you're teaching physics or curious about real-world motion, this example shows how gravity governs vertical trajectories — from a simple throw to a dramatic landing.", "---", "Keywords: projectile motion, time to fall, initial velocity 20 m/s, height 50 m, ground impact time, physics equations, quantum mechanics misconception (not related), gravity 9.8 m/s², kinematics, quadratic equation, sphere motion, upward throw calculation.", "---", "Optimization Tips for SEO:\n- Use clear, conversational headings\n- Include relevant keywords naturally\n- Break down complex problems into digestible steps\n- Add real-world context to enhance user engagement and visibility", "---", "Table: Key Values Recap", "| Parameter | Value |\n|-------------------|-------------------|\n| Initial velocity ((u)) | 20 m/s |\n| Initial height ((s_0)) | 50 meters |\n| Acceleration ((a)) | -9.8 m/s² |\n| Equation solved | ( 4.9t^2 - 20t - 50 = 0 ) |\n| Time to ground | ≈ 5.83 seconds |", "---", "Understanding motion isn’t just about numbers — it’s about seeing physics in every jump, throw, and fall."]