A 76.0 kg boulder is rolling horizontally at the top of a vertical cliff that is

Question

A 76.0 kg boulder is rolling horizontally at the top of a vertical cliff that is 20.0 m above the surface of a lake, as shown in figure below. 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 25.0 m below the top of the dam.

A. What must the minimum speed of the rock be just as it leaves the cliff so that it will travel to the plain without striking the dam? (Answer in m/s)

B. How far from the foot of the dam does the rock hit the plain? (Answer in meters)

Explanation / Answer

Fin the time that it would take the boulder to drop the vertical 20m using vertical velocity. s = ut + 1/2 at^2 20 = 0 + 1/2 x 9.8 t^2 (note u = 0 as no initial vertical velocity) t = 2.02sec For the boulder to travel 100, in 2.02sec, means Vh = 100/2.02 = 49.5m/s b) using the same philosophy, find the time for the boulder to hit the plain s = ut+1/2 at^2 45 = 0 + 1/2 x 9.8 t^2 t = 3.03secs The horizontal velocity of the boulder is 49.5m/s therefore distance travelled = 49.5 x 3.03 = 150m. The boulder therefore lands 150 - 100 = 50m from the dam wall

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