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Projectile Motion Physics Problems

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+ ward ' s science 5100 West Henrietta Road • PO Box 92912 • Rochester, New York 14692-9012 • p: 800 962-2660 • wardsci.com Find materials for this activity at wardsci.com. Discover more free activities at wardsworld.wardsci.com Example Problems: 2) What is the total time in flight of the rocket? The formulas we need to solve this problem are the following: • u x = u0 • cosθ • u y = u0 • sinθ – (g • t) t at the time of launch is t1 which = 0 t at the time the rocket lands (ground) is t2. At t2, y = 0. Therefore, if y = u • t • sinθ – 0.5gt2: y = t(u • sinθ – 0.5gt) or y = 2(u • sinθ) – gt At t2, y = 0, so 2(u • sinθ) – gt = 0 t = 2(u • sinθ) / g t2 = 2(25 • sin 30) / 9.8 t2 = 2.55 sec so time of flight = t2 – t1 = 2.55 – 0 = 2.55 seconds 3) What is the horizontal range (distance covered) of the rocket? The formulas we need to solve this problem are the following: • x = u • t • cosθ So in this case, we are looking for x at t2 – t1 (which is 2.55 seconds from part b) x = 25 • 2.55 • cos 30 = 32.3 meters 4) What is the magnitude of the velocity of the object just before it hits the ground? The formulas we need to solve this problem are the following: • u x = u0 • cosθ • u y = u0 • sinθ − (g • t) • u = √(u2 x + u2 y) We are solving for the velocity components at t2 = 2.55 seconds u x = u0 • cosθ u x = 25 (cos 30) = 23.5 m/s u y = u0 • sinθ − (g • t) u y = 25 (sin 30) – (9.8 • 2.55) = - 12.5 m/s u = √(u2 x + u2 y) u = √(23.5)2 + (-12.5)2 = 26.6 m/s

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