A ball is thrown upward with initial velocity 32 m/s from 20 meters high. When does it hit the ground? (Use \( g = 9.8 \, \text{m/s}^2 \))

["# How Long Does a Ball Stay in the Air When Thrown Upward? Solving a Real-World Physics Problem", "When a ball is thrown upward with an initial velocity, physics enthusiasts and students alike often wonder: When does the ball hit the ground? This article explains how to calculate the time from release to impact using basic kinematics, with real numbers and practical application.", "### Problem Setup", "A ball is thrown upward with:\n- Initial velocity ( v_0 = 32 , \ ext{m/s} )\n- Launch height ( h_0 = 20 , \ ext{m} )\n- Acceleration due to gravity ( g = 9.8 , \ ext{m/s}^2 ) (acting downward)", "We need to determine when the ball hits the ground — that is, the time ( t ) from launch until its height reaches zero meters.", "---", "## Step-by-Step Kinematic Solution", "### 1. Write the vertical motion equation", "The vertical displacement ( y(t) ) at any time ( t ) is given by:", "[\ny(t) = h_0 + v_0 t - \frac{1}{2} g t^2\n]", "We want to find the time ( t ) when the ball hits the ground, i.e., when ( y(t) = 0 ):", "[\n0 = 20 + 32t - \frac{1}{2}(9.8)t^2\n]", "Simplify the equation:", "[\n0 = 20 + 32t - 4.9t^2\n]", "Rewriting in standard quadratic form:", "[\n4.9t^2 - 32t - 20 = 0\n]", "This is a quadratic equation of the form ( at^2 + bt + c = 0 ), with:\n- ( a = 4.9 )\n- ( b = -32 )\n- ( c = -20 )", "---", "### 2. Apply the quadratic formula", "The solutions are given by:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "First compute the discriminant:", "[\n\Delta = b^2 - 4ac = (-32)^2 - 4(4.9)(-20) = 1024 + 392 = 1416\n]", "Now compute the square root:", "[\n\sqrt{1416} \approx 37.63\n]", "Now plug into the formula:", "[\nt = \frac{32 \pm 37.63}{2 \cdot 4.9} = \frac{32 \pm 37.63}{9.8}\n]", "This gives two solutions:", "1. ( t = \frac{32 + 37.63}{9.8} = \frac{69.63}{9.8} \approx 7.11 , \ ext{s} )\n2. ( t = \frac{32 - 37.63}{9.8} = \frac{-5.63}{9.8} \approx -0.57 , \ ext{s} )", "---", "### 3. Select the physically meaningful solution", "Time cannot be negative, so we discard the negative root.", "Thus, the ball hits the ground at approximately:", "[\nt \approx 7.11 , \ ext{seconds}\n]", "---", "## Why This Even Matters (Application & Takeaways)", "Understanding how long an object stays in the air under gravity helps in sports, engineering, and safety calculations. In projectile motion, knowing the flight time ensures accurate predictions and timing — whether launching a ball, designing a jump, or analyzing trajectories.", "In this case, the initial upward velocity of 32 m/s carries the ball high, but gravity pulls it back down, with the 20 m height adding to the duration before hitting the ground.", "---", "## Quick Summary Table", "| Quantity | Value |\n|-----------------------|------------------------|\n| Initial height ( h_0 )| 20 meters |\n| Initial velocity ( v_0 )| 32 m/s (upward) |\n| Gravity ( g ) | 9.8 m/s² (downward) |\n| Time to ground ( t ) | ~7.11 seconds |", "---", "## Conclusion", "By applying the kinematic equation for vertical motion under constant acceleration, we determined that a ball thrown upward with 32 m/s from a 20 m height hits the ground after approximately 7.11 seconds. This straightforward but powerful method illustrates how physics models real-world phenomena with precision.", "If you’re solving similar problems, always remember:\n- Apply the correct sign conventions (upward is positive).\n- Use ( g ) as a downward acceleration.\n- Use the quadratic formula to solve for time when displacement is zero.", "For further reading on projectile motion and time of flight, visit Khan Academy Physics or consult kinematics textbooks."]