Consider a rock thrown off a bridge of height 78.9 m at an angle = 25° with resp
ID: 584211 • Letter: C
Question
Consider a rock thrown off a bridge of height 78.9 m at an angle = 25° with respect to the horizontal as shown in Figure P4.20. The initial speed of the rock is 10.2 m/s. Find the following quantities:
(a) the maximum height reached by the rock
(b) the time it takes the rock to reach its maximum height
(c) the time at which the rock lands
(d) how far horizontally from the bridge the rock lands
(e) the velocity of the rock (magnitude and direction) just before it lands.
magnitude
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!)
Explanation / Answer
(a) Let the rock attain h meter in t1 sec
=>By v^2 = u^2 - 2gh
=>0 = [Uy]^2 - 2gh
=>h = [10.2 x sin25*]^2/[2 x 9.8]
=>h = 0.948 m
(b) Let the rock attain its maximum height in t1 sec
=>By v = u - gt
=>0 = [Uy] - gt1
=>t1 = usin/g
=>t1 = [10.2 x sin25*]/9.8
=>t1 0.44 sec
(c) Total height (H) of the stone fro ground = h + 78.9 = 79.848 m
Let the stone take t2 sec to fall H meter
=>By s = ut + 1/2gt^2
=>79.848 = 0 + 1/2 x 9.8 x t2^2
=>t2 = 16.29
=>t2 4.03 sec
Thus the total time to land (T) = t1 + t2 = 0.44 + 4.03 = 4.47 sec
(d) By R = [Ux] x T
=>R = ucos x T
=>R = 10.2 x cos25* x4.47
=>R = 41.32 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 4.03 = 39.49 m/s
& Vx = Ux = 10.96 m/s
=>v = [(Vx)^2 + (Vy)^2]
=>v = [(10.96)^2 + (39.49)^2]
=>v = 40.98 m/s
By tan = Vy/Vx = 39.49/10.96 = 74 degree