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Projectile Motion Physics Problems

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+ ward ' s science Maximum Height: The maximum height is reached when u y = 0. Using this we can rearrange the velocity equation to find the time it will take for the object to reach maximum height t h = (u • sinθ) / g where t h stands for the time it takes to reach maximum height. From the displacement equation we can find the maximum height h = (u2 • sin2θ) / (2 • g) 0 Y X Example Problems: A small rocket is launched at a 30 degree angle at 25 m/s. 1) What is the maximum height reached by the rocket? The formulas we need to solve this problem are the following: • u x = u0 • cosθ • u y = u0 • sinθ – (g • t) • g = 9.8 m/s2 At its maximum height, u y = u0 • sinθ – (g • t) = 0 First, solving for t: t = (u0 • sinθ) / g t = (25 • sin 30) / 9.8 t = 1.28 seconds Now solve for y: y = u • t • sinθ – 0.5gt2 y = (25 • 1.28 • sin 30) – (0.5 • 9.8 • 1.282) y (max) = 7.97 m The Science of Projectile Motion θ Range (R) θ Maximum Height (H) u x = u0 • cosθ u y = u sinθ u u x = u cosθ u u x = component of velocity along x – axis, a x = acceleration along x – axis = 0 u y = component of velocity along x – axis, ay = acceleration along y – axis = -g

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