Consider a rock that is thrown off a bridge of height 85 m at an angle ? = 23 Co

Question

Consider a rock that is thrown off a bridge of height 85 m at an angle ? = 23

Consider a rock that is thrown off a bridge of height 85 m at an angle ? = 23° with respect to the horizontal as shown in the figure below. If the initial speed the rock is thrown is 11 m/s, find the following quantities. (a) The time it takes the rock to reach its maximum height. _______s (b) The maximum height reached by the rock. ______ m (c) The time at which the rock lands. ______ s (d) The place where the rock lands. ______ m (e) The velocity of the rock (magnitude and direction) just before it lands. magnitude_____ m/s direction____ °

Explanation / Answer

(a) Let the rock attain its maximum height in t1 sec
=>By v = u - gt
=>0 = [Uy] - gt1
=>t1 = usin?/g
=>t1 = [11 x sin23*]/9.8
=>t1 ? 0.438 sec
(b) Let the rock attain h meter in t1 sec
=>By v^2 = u^2 - 2gh
=>0 = [Uy]^2 - 2gh
=>h = [11 x sin23*]^2/[2 x 9.8]
=>h = 0.942 m
(c) Total height (H) of the stone fro ground = h + 85 = 85.942 m
Let the stone take t2 sec to fall H meter
=>By s = ut + 1/2gt^2
=>85.942 = 0 + 1/2 x 9.8 x t2^2
=>t2 = ?17.539
=>t2 ? 4.187sec
Thus the total time to land (T) = t1 + t2 = 0.438 + 4.187 = 4.625 sec
(d) By R = [Ux] x T
=>R = ucos? x T
=>R = 11 x cos23* x 4.625
=>R = 46.83 m
(e) Let the angle of landing is ? and the magnitude of velocity is v m/s
=>By v = u + gt
=>Vy = 0 + 9.8 x t2
=>Vy = 9.8 x 4.187 = 41.03 m/s
& Vx = Ux = 4.434 m/s
=>v = ?[(Vx)^2 + (Vy)^2]
=>v = ?[(4.434)^2 + (41.03)^2]
=>v = 41.26 m/s
By tan? = Vy/Vx = 41.03/4.434 = 9.25 ? tan83.8*
=>? = 83.8*

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