Consider a rock that is thrown off a bridge of height 81 m at an angle ? = 23 Co

Question

Consider a rock that is thrown off a bridge of height 81 m at an angle ? = 23

Consider a rock that is thrown off a bridge of height 81 m at an angle ? = 23 degree with respect to the horizontal as shown in the figure below. If the initial speed the rock is thrown is 12 m/s, find the following quantities. The time it takes the rock to reach its maximum height. The maximum height reached by the rock. The time at which the rock lands. The place where the rock lands. he velocity of the rock (magnitude and direction) just before it lands.

Explanation / Answer

(a) Let the rock attain its maximum height in t1 sec
=>By v = u - gt
=>0 = [Uy] - gt1
=>t1 = usin theta /g
=>t1 = [12 x sin23*]/9.8
=>t1 = 0.47 sec

(b) Let the rock attain h meter in t1 sec
=>By v^2 = u^2 - 2gh
=>0 = [Uy]^2 - 2gh
=>h = [12 x sin23*]^2/[2 x 9.8]
=>h = 1.11 m
Total height (H) of the stone fro ground = h + 81 = 82.11 m

(c)Let the stone take t2 sec to fall H meter
=>By s = ut + 1/2gt^2
=>82.11 = 0 + 1/2 x 9.8 x t2^2
=>t2 = sqrt(16.75)
=>t2 = 4.09sec
Thus the total time to land (T) = t1 + t2 = 0.47 + 4.09 = 4.56 sec

(d) By R = [Ux] x T
=>R = ucos? x T
=>R = 12 x cos23* x 4.56
=>R = 50.37 m

(e) Let the angle of landing is alpha(a) and the magnitude of velocity is v m/s
=>By v = u + gt
=>Vy = 0 + 9.8 x t2
=>Vy = 9.8 x 4.09 = 40.08 m/s
& Vx = Ux =cos 23 * 12 = 11.04 m/s
=>v = sqrt[(Vx)^2 + (Vy)^2]
=>v = sqrt[(11.04)^2 + (40.08)^2]
=>v = 41.57 m/s
By tan a = Vy/Vx = 40.08/11.04 = 3.63
=>a = 74.5 degree

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