Question

Consider a rock that is thrown off a bridge of height 68 m at an angle ? = 22

Answer 1

a) at max height Vy = 0

so , using Vy = Vyo - gt

=> 0 = 12 sin22 - 9.8t

=> t = 12sin22 / 9.8 = 0.458 = 0.46 s (approx)

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b) using Vy^2 = Vyo^2 - 2gh u get :

0 = (12 sin22)^2 - 2 (9.8) h

solving for h u get h = (12 sin22)^2 / 2(9.8) = 1.03 m

adding that to 22m ( the bridge's height) u get :

h max = 68 + 1.03 = 69.03 m

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c) when the rock lands then h = - 68 m

so - 68 = 12sin22 t - 0.5(9.8) t^2

=> -68 = 4.49 t - 4.9 t^2

=> 4.9 t^2 - 4.49 t - 68 = 0

t = [4.49 + sqrt (4.49^2 + 4(4.9)(68))] / 2(4.9) = 4.211 s

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d) Vx remains consant , so we use Vx = x / t

=> x = Vx t = 12 cos22 x 4.211 = 46.85 m away from the bridge

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e) Vx = 12 cos 22 = 11.12 m/s

now find Vy :

Vy = 12sin22 - 9.8 (4.211) = - 36.77 m/s

so V = sqrt(11.12^2 + 36.77^2) = 38.41 m/s

angle = arctan (-36.77/ 11.12) = - 73.17 degrees