A 60.0- k g boulder is rolling horizontally at the top of a vertical cliff that

Question

A 60.0-kg boulder is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake, as shown in the figure (Figure 1) . The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25 m below the top of the dam.

A) What must be the minimum speed of the rock just as it leaves the cliff so it will travel to the plain without striking the dam?

B) How far from the foot of the dam does the rock hit the plain?

A 60.0-kg boulder is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake, as shown in the figure (Figure 1) . The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25 m below the top of the dam. What must be the minimum speed of the rock just as it leaves the cliff so it will travel to the plain without striking the dam? How far from the foot of the dam does the rock hit the plain?

Explanation / Answer

A )H = ut + at^2/2

20 = 0 + 9.8t^2/2

t = 2.02 sec

R = v_xt   = vo x 2.02 = 100

vo = 49.50 m/s

B)

vertical component at dam v_y = 0 - 9.8 x2.02 = 19.80 m/s

so 25 = 19.80t + 9.8t^2/2

4.9t^2 + 19.80t - 25 = 0

t = 1.01 sec

x = vot = 49.50 x 1.01 = 50 m

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