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with what i know the ball is thown at a pace of 45.0m and lands 24.0m from base so lets draw the big 5 diagrams and put them together and then we find time because its what needed to evaluate the position the ball is in.

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12y ago
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14y ago

Ok let's solve this doing it the algebra way then the calculus way. Finding TIME: Given: acceleration = -9.81 m/s-2 initial velocity = 25 m/s final velocity (at the top of the trajectory, the highest elevation) = 0 time = ? Formula: a = Δv/Δt Rearranged: Δt = Δv/a = -25/-9.81 = 2.54842 seconds (multiply by two since this is the time taken to reach the "top" of the trajectory) So t = 5.09684 s = 5.1 seconds (2 sigdigs) Finding MAXIMUM HEIGHT: Let's use d = vft - 0.5at2 NOTE: use half of the time of 5.09684 seconds since it will take half this time to reach the MAX HEIGHT d = (0) - 0.5(-9.81)(2.54842)2 = 32m ---- Calculus way: Finding TIME: a = -9.81 Antiderivative: v = -9.81t + 25 Antiderivative: d = -4.905t2 + 25t (solve for t when d = 0 (the projectile is on the ground)) t = 0 s, 5.1 s (omit 0 since this is the starting time when d = 0) t = 5.1 seconds Finding MAXIMUM HEIGHT When v = 0, the ball will be at max height. Sub this t-value into the distance formula. v = -9.81t +25 = 0 t = 2.5484 s d = -4.905(2.5484)2 + 25(2.5484) = 32 m

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12y ago

9.17 secs

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Q: A ball is thrown vertically upward from the ground at a speed of 45 ms How long does it take to return to the ground?
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