Problem 3.63 A 76.0kg boulder is rolling horizontally at the top of a vertical c

Question

Problem 3.63

A 76.0kg boulder is rolling horizontally at the top of a vertical cliff that is 20.0m above the surface of a lake, as shown in figure below.(Figure 1) The top of the vertical face of a dam is located 100m 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 25.0m below the top of the dam.

Part 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?

Part B

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

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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