HW-4-solutions - pandey(sp34566 HW-4 swinney(57305 This...

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pandey (sp34566) – HW-4 – swinney – (57305) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points Consider the motion of a projectile fired at t = 0 with initial velocity v 0 at an angle θ from the horizontal. x y v 0 θ h R What is the height of the trajectory? t top = v 0 y g , so the maximum height is h = h 0 + v 0 y t top 1 2 g ( t top ) 2 = 0 + v 0 y v 0 y g 1 2 g parenleftbigg v 0 y g parenrightbigg 2 = v 2 0 y 2 g . 002(part2of2)10.0points Consider two cases with different initial angles θ 1 and θ 2 and the same initial speed v 0 . x y v 0 θ 2 v 0 θ 1
pandey (sp34566) – HW-4 – swinney – (57305) 2 t 2 t 1 = 2 v 0 sin θ 2 g 2 v 0 sin θ 1 g = sin θ 2 sin θ 1 . 003(part1of3)10.0points A soccer ball is kicked at an angle of 62 to the horizontal, with an initial speed of 22 m / s. Assume for the moment that we can neglect air resistance. g = 9 . 8 m / s 2 . How long is the ball in the air?
Explanation: To determine how long the ball is in the air, we can use the position update formula derived from the momentum principle: y f = y i + v i,y Δ t + 1 2 a y Δ t 2 . We just need to fill in the quantities that we know. Clearly, y f = y i = 0 m and a y = g = 9 . 8 m / s 2 . The only other thing we need to solve for the time is the initial velocity in the y direction. We obtain this from trigonometry: v i,y = v sin 62 = (22 m / s) sin 62 = 19 . 4248 m / s . So, rearranging the above equation to solve for the time, we obtain 0 = v i,y Δ t + 1 2 a y Δ t 2 v i,y Δ t = 1 2 a y Δ t 2 Δ t = 2 v i,y a y = 2 (19 . 4248 m / s) 9 . 8 m / s 2 = 3 . 96425 s . 004(part2of3)10.0points How far does the ball go (horizontal distance along the field)?

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