jungkook whose height is 1.17m,threw the frisbee towards his dog w/an initial horizontal velocity of 8.0m/s how far was the dog from jungkook
Answer (1)
Answer:The dog was approximately 3.91 meters away from Jungkook.Find the Time of FlightThe frisbee falls freely under gravity from a height of 1.17 m, so we use the free-fall equation:[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{y = \frac{1}{2} g t^2}[/tex]where:y = 1.17 m (initial height)g = 9.81 m/s² (acceleration due to gravity)t = time of flightSolving for t:[tex] \tt{t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2(1.17)}{9.81}}}[/tex][tex]{t = \sqrt{\frac{2.34}{9.81}} = \sqrt{0.2385}}[/tex][tex]t \approx 0.4884 \text{ s}[/tex]Find the Horizontal DistanceSince horizontal velocity remains constant at 8.0 m/s, the horizontal distance is:[tex] \tt{x = v t}[/tex][tex]x = (8.0)(0.4884)[/tex][tex]x \approx 3.91 \text{ m}[/tex]
