What is the minimum speed of the rock as it leaves the cliff if 76.0 rm kg boulder is rolling horizontally at a vertical cliff's top 20.0 rm m above a lake's surface SEE Discuss?

Answer 1

As given it is impossible to answer the question; the shape of the dam is missing.

The shape of the dam is important, because the boulder may just miss the top of the dam but hit part of the dam lower down: dams are not usually built as a straight vertical wall as the horizontal force of the water needs to be counteracted - they are usually built as an arc, and with the foot of the dam wider than the top so that they slope away from the top and away from the water.

At a guess, the figure shows a simple dam which is vertical, and I shall assume as such for the calculation, but give a note how to fix for a sloped dam with a wider base than top that the boulder totally misses.

I assume that air resistance is to be ignored (due to the massiveness of the boulder).

Throughout the motion of the boulder through the air to the valley floor there is only one (significant) force acting on the boulder: gravity. Thus the horizontal velocity of the boulder will not (significantly) change, but the vertical velocity will increase.

This requires use of an equation of motion on the separate components of the velocity of the boulder:

s = ut + ½at²

a) To calculate the horizontal velocity requires the time is given by the time to drop 20m under gravity, and the distance is 100 m; thus:

For the drop:

u = 0

a = g ≈ 9.81 m/s²

s = 20 m

→ 20 m = 0 + ½ × g × t²

→ t² = 20 m × 2 ÷ g

→ t = √(20 m × 2 ÷ g) = √(40m/g) ≈ 2.02 s

For the horizontal movement:

t = √(40m/g) s

a = 0 m/s²

s = 100 m

→ 100 m = u × √(40m/g) + 0

→ u = 100 m ÷ √(40m/g) ≈ 49.5 m/s

b) To calculate how far from the base of the dam requires the time to drop from the top of the cliff to the valley under gravity, the speed calculated in part (a), and the distance of the base of the dam from the base of the cliff; thus:

For the drop:

u = 0

a = g ≈ 9.81 m/s²

s = 20 m + 25 m = 45 m

→ 45 m = 0 + ½ × g × t²

→ t² = 45 m × 2 ÷ g

→ t = √(45 m × 2 ÷ g) =√(90m/g) ≈ 3.03 s

For the horizontal movement:

t = √(90m/g) s

a = 0 m/s²

u = 100 m ÷ √(40m/g) m/s

→ s = 100 m ÷ √(40m/g) m/s × √(90m/g) s + 0

→ s = 100 m × √(90/40)

→ s = 100 m × 3/2 = 150 m

As the top of the dam is 100 m from the base of the cliff, it has travelled a further 150 m - 100 m = 50 m relative to the top of the dam,

Thus it will land 50 m from the base of the dam assuming the dam is vertical.

IF the dam is not vertical but (sort of) triangular with the base wider than the top, then subtract the extra width of the base (away form the cliff) from the 50 m. for example if the base of the dam is 10 m wide, ie the base is 10 m further from the cliff than the top, then the boulder would land 50 m - 10 m = 40 m from the base of the dam.

Note that the mass of the boulder is irrelevant as it does not come into the equations, except to indicate that air resistance is negligible and can be ignored in the calculations. For a much lighter projectile (or one with large drag), there will be a significant horizontal force on it which will produce a significant negative acceleration which will need to be taken into account.