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1/ . ill I lqwen: 12 = 13:,m = 0.5 kg,a = 37,h, = 2m,g = 9.8:, Initial vertical velocity: vyo = 17. sin37 Horizontal
1/ . ill "I lqwen: 12 = 13:,m = 0.5 kg,a = 37,h, = 2m,g = 9.8:, Initial vertical velocity: vyo = 17. sin37 Horizontal velocity of the ball throughout its ight: 17,0 = 17. cos 37 -) Height ofthe ball at time twill be: h = gt2 + vyot The ball is caught on its way down at 2m: h = hf > 2 = $9.8t2 + 13 x sin(37) t - 4.9:2 7.82t + 2 = o 391 i V54881 490 - t = 1.28 (s) or t = 0.32 (s) -) These two t values correspond to the the ball being at 2m on its way up and on its way down Because the ball was on its way down after reaching maximum height, we should pick the higher t value which corresponds to a later time t = 1.28 s Horizontal distance the ball has travelled: d = 17,0. t = 11 cos(37) t = 13 cos(37).1.28 = 8.81 (m) )t=
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