Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands three times as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.

3 : 1

Explanation:

Given that One stone lands three times as far from the base of the building from which it was thrown as does the other stone.

To Find the ratio of the height of the taller building to the height of the shorter building, let consider their range and maximum height

Maximum height = H = u²sin²∅/2g

Horizontal range = R = u²sin2∅/g

1]

H = u²sin²∅/2g

2H/sin²∅ = u²/g (1)

2]

R = u²sin2∅/g

R/sin2∅ = u²/g (2)

From equation (1) and (2)

2H/sin²∅ = R/sin2∅

2H/sin∅×sin∅ = R/2sin∅cos∅

2H/sin∅ = R/2cos∅

2H × 2cos∅ = R × sin∅

4Hcos∅ = Rsin∅

R = 4Hcos∅/sin∅

[ R = 4H × cot∅ ]

Since One stone lands three times as far from the base of the building from which it was thrown as does the other stone. 

That is, R1 = 3R2

Shorter building = 4H × cot∅

Higher building = 3(4H × cot∅)

Ratio = 12H × cot∅ / 4H × cot∅

Ratio = 12H/4H

Ratio = 3

33: i think it’s b

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