Consider a rock thrown off a bridge of height 76 m at an angle = 25° with respec

Question

Consider a rock thrown off a bridge of height 76 m at an angle = 25° with respect to the horizontal as sh (a) the maximum height reached by the rock 78.54 m n Figure P4.20. The initial speed of the rock is 16.7 m/s. Find the following quantitie Bridge 25° You are correct. ie Your receipt no. is 167-1858 (b) the time it takes the rock to reach its maximum height 0.720 s oc You are correct vi Your receipt no. is 167-82450) (c) the time at which the rock lands Submit Answer Tries 0/10 (d) how far horizontally from the bridge the rock lands Submit Answer Tries 0/10 (e) the velocity of the rock (magnitude and direction) just before it lands magnitude Submit Answer Tries 0/10 direction: Give your answer in degrees. If you think the answer is "20 degrees down from the positive x-axis", you would enter "-20" (note the negative sign!) Submit Answer Tries 0/10 Post Discussion Send Feedback

Explanation / Answer

a) using Vy^2 = Vyo^2 - 2gh u get :
0 = (16.7 sin25)^2 - 2 (9.8) h
solving for h u get h = (16.7 sin25)^2 / 2(9.8) = 2.541 m
adding that to 76 m ( the bridge's height) u get :
h max = 76 + 2.541 = 78.541 m

b) at max height Vy = 0
so , using Vy = Vyo - gt
=> 0 = 16.7sin25 - 9.8t
=> t = 16.7sin25 / 9.8 = 0.72 s (approx)

c) when the rock lands then h = - 76 m
so - 76 = 16.7sin25* t - 0.5(9.8) t^2
=> -76 = 7.058 t - 4.9 t^2
=> 4.9 t^2 - 7.058 t - 76 = 0
t = 4.72 s

d) Vx remains consant , so we use Vx = x / t
=> x = Vx t = 16.7 cos25 x 4.72 = 71.44 m away from the bridge

e) Vx = 16.7 cos 25 = 15.14 m/s
now find Vy :
Vy = 16.7sin25 - 9.8 (4.72) = -39.2 m/s
so V = sqrt(15.14^2 + 39.2^2) = 42.02 m/s
angle = arctan (-39.2 / 15.14) = - 68.9 degrees

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