Working on test corrections and I think I figured out the answer to number 8, bu

Question

Working on test corrections and I think I figured out the answer to number 8, but I’m not sure how to get the height of the cliff (#9) 20m X 20 11.55 (eo-ad hyt |For problems 8 & 9, refer to the following on-OP information ad ul Your friend throws a rock up from a cliff 1.2, tic 9.B ml a at an angle of 60 degrees. You observe it to go 20 m vertically before beginning to fall back down. The time it takes to reach the highest point is 1.2 seconds.(Acceleration due to gravity is 9.8 m/s) Hint: Use a trig relation to connect the initial velocities horizontally and verticallv.l If the rock lands 45 m from the cliff base, how long was the rock in flight? a. ox sec e. 17. sec Vo ton O How high is the elifr? d. 256 m e. 457m = 11.76 45 1.79

Explanation / Answer

a)

Vo = initial velocity of launch

consider the motion from the point of launch to maximum height along the vertical direction

Vfy = final velocity at the maximum height = 0

Voy = initial velocity = Vo Sin60

a = acceleration = - 9.8

t = time = 1.2 sec

using the equation

Vfy = Voy + at

0 = Vo Sin60 + (- 9.8) (1.2)

Vo = 13.6 m/s

consider the motion along the horizontal direction

X = horizontal displacement = 45 m

Vox = constant velocity along horizontal direction = Vo Cos60 = (13.6) Cos60 = 6.8 m/s

t' = time of travel

using the equation

t' = X/Vox = 45/6.8 = 6.62 sec

b)

Yo = initial position = height of cliff = h

Y = final position = 0

using the equation

Y = Yo + Voy t' + (0.5) a t'2

0 = h + (13.6) Sin60 (6.6) + (0.5) (-9.8) (6.6)2

h = 135.71 m

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