A ball is thrown upward from a 45-meter tall building with initial velocity 20 m/s. Its height is modeled by \( h(t) = -5t^2 + 20t + 45 \). When does it hit the ground?

["Title: When Does a Ball Hit the Ground When Thrown Upward from a 45-Meter Building? | Physics Model Using ( h(t) = -5t^2 + 20t + 45 )", "---", "When a ball is thrown upward from a 45-meter tall building with an initial velocity of 20 m/s, its motion follows a quadratic trajectory governed by the height function:\n[ h(t) = -5t^2 + 20t + 45 ]\nThis equation models the ball’s height ( h(t) ) (in meters) as a function of time ( t ) (in seconds). Understanding when the ball hits the ground involves solving for ( t ) when ( h(t) = 0 ).", "### The Physics Behind the Equation\nThe coefficient of ( t^2 ) in the quadratic formula (( -5 )) represents the effect of gravity,约(-9.8, \ ext{m/s}^2), pulling the ball downward. The linear term ( 20t ) captures the initial upward velocity, while the constant ( 45 ) indicates the starting height.", "### Solving for Time When the Ball Hits the Ground\nTo find when the ball lands, set ( h(t) = 0 ):\n[\n-5t^2 + 20t + 45 = 0\n]\nDividing the entire equation by (-5) simplifies the equation:\n[\nt^2 - 4t - 9 = 0\n]", "Apply the quadratic formula ( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -4 ), and ( c = -9 ):\n[\nt = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-9)}}{2(1)} = \frac{4 \pm \sqrt{16 + 36}}{2} = \frac{4 \pm \sqrt{52}}{2}\n]\nSimplify ( \sqrt{52} = 2\sqrt{13} ), so:\n[\nt = \frac{4 \pm 2\sqrt{13}}{2} = 2 \pm \sqrt{13}\n]", "Since time cannot be negative, only the positive root is physically meaningful:\n[\nt = 2 + \sqrt{13}\n]\nApproximating ( \sqrt{13} \approx 3.605 ), we get:\n[\nt \approx 2 + 3.605 = 5.605 \ ext{ seconds}\n]", "### Final Answer\nThe ball hits the ground after approximately ( 2 + \sqrt{13} ) seconds, which is about 5.6 seconds based on decimal estimation. This calculation precisely determines the time using the quadratic height model.", "---", "Summary:\n- The ball is thrown from a 45 m tall building with a 20 m/s upward velocity.\n- Its height modeled by ( h(t) = -5t^2 + 20t + 45 )\n- The ball hits the ground when ( h(t) = 0 ), solved via the quadratic equation.\n- Exact time: ( t = 2 + \sqrt{13} ) seconds; approximately 5.6 seconds.", "Keywords: ball thrown upward, quadratic height model, physics classroom, projectile motion, time when ball hits ground, ( h(t) = -5t^2 + 20t + 45 ), ( t = 2 + \sqrt{13} ", "---", "Follow this precise method to analyze motion in similar physics problems. Understand the model, solve the quadratic, and always check for physically valid solutions!"]