Jeff throws a baseball up in the air. His friends want to know how long it takes for the baseball to hit the ground if he throws it at 50 mph, 10 feet high.
Answer (1)
[tex]\begin{gathered}\begin{gathered}{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} \\\end{gathered}\end{gathered}[/tex] To determine how long it takes for a baseball thrown upward at 50 mph from a height of 10 feet to hit the ground, we can use the principles of kinematics. Here’s how we can break it down step-by-step: Convert Units First, we need to convert the speed from miles per hour (mph) to feet per second (ft/s), since the height is given in feet. 1 mile = 5280 feet1 hour = 3600 seconds To convert 50 mph to feet per second: [tex]50 \text{ mph} = 50 \times \frac{5280 \text{ feet}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} \approx 73.33 \text{ ft/s}[/tex] Use Kinematic Equation Now we can use the kinematic equation for motion under gravity to find the time it takes for the baseball to hit the ground: [tex]h = v_0t + \frac{1}{2}at^2[/tex] Where: h is the height (which will be 0 when it hits the ground) [tex]v_0[/tex] is the initial velocity (73.33 ft/s) a is the acceleration due to gravity (approximately -32.2 ft/s², negative because it acts downward)t is the time in seconds Set Up the Equation Since we start at 10 feet high and end at 0 feet when it hits the ground, we can set up the equation: [tex]0 = 10 + 73.33t - \frac{1}{2}(32.2)t^2[/tex] This simplifies to: [tex]0 = 10 + 73.33t - 16.1t^2[/tex] Rearranging the Equation Rearranging gives us a quadratic equation: [tex]16.1t^2 - 73.33t - 10 = 0[/tex] Solve the Quadratic Equation We can use the quadratic formula: [tex]t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex] Where a = 16.1, b = -73.33, and c = -10. Calculating the discriminant: [tex]b^2 - 4ac = (-73.33)^2 - 4 \cdot 16.1 \cdot (-10) \approx 5377.7 + 644 = 6021.7[/tex] Now applying the quadratic formula: [tex]t = \frac{-(-73.33) \pm \sqrt{6021.7}}{2 \cdot 16.1}[/tex] Calculating the values: [tex]\sqrt{6021.7} \approx 77.57[/tex] Thus, [tex]t = \frac{73.33 \pm 77.57}{32.2}[/tex] This gives us two potential times: [tex]t_1 = \frac{150.9}{32.2} \approx 4.68 \text{ seconds (valid time)}[/tex][tex]t_2 = \frac{-4.24}{32.2} \text{ (not valid, as time can’t be negative)}[/tex] Conclusion The baseball will take approximately 4.68 seconds to hit the ground after being thrown upwards at 50 mph from a height of 10 feet. [tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\begin{gathered} \boxed{\begin{array}{cc} \sf \footnotesize \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ···\: ʚ\: \: \: \: \: \: Hope\:it\:helps \: \: \: \: \: ɞ \:··· \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\footnotesize\:\# CarryOnLearning \\ \sf\footnotesize ૮₍´˶ \: • \: . \: • \: ⑅ ₎ა \: \leadsto \footnotesize\sf\color{purple} \underline{Study\:Well!}\end{array}}\end{gathered}[/tex][tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\large\qquad\qquad\qquad\tt MARCH/13/2025 [/tex]
