Consider a rock that is thrown off a bridge of height 26 m at an angle = 21 Solu

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change in values plug your values and make calculations a) at max height Vy = 0 so , using Vy = Vyo - gt => 0 = 15sin21 - 9.8t => t = 15sin21 / 9.8 = 0.548 = 0.55 s (approx) ***************************************… b) using Vy^2 = Vyo^2 - 2gh u get : 0 = (15 sin21)^2 - 2 (9.8) h solving for h u get h = (15 sin21)^2 / 2(9.8) = 1.47 m adding that to 22m ( the bridge's height) u get : h max = 22 + 1.47 = 23.47 m ***************************************… c) when the rock lands then h = - 22 m so - 22 = 15sin21 t - 0.5(9.8) t^2 => -22 = 5.38 t - 4.9 t^2 => 4.9 t^2 - 5.38 t - 22 = 0 t = [5.38 + sqrt (5.38^2 + 4(4.9)(22))] / 2(4.9) = 2.74 s ***************************************… d) Vx remains consant , so we use Vx = x / t => x = Vx t = 15 cos21 x 2.74 = 38.37 m away from the bridge ***************************************… e) Vx = 15 cos 21 = 14 m/s now find Vy : Vy = 15sin21 - 9.8 (2.74) = - 21.48 m/s so V = sqrt(14^2 + 21.48^2) = 25.64 m/s angle = arctan (-21.48 / 14) = - 56.9 degrees

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