Question

The distance at which nile can throw a javelin is inversely proportional to the weight of that javelin. Nile can throw a 1.5-pound javelin exactly 260 feet. How far would nile be able to throw a 2-pound javelin, in feet, assuming all other factors remain constant? (disregard units

Answer 1

Answer---> 195 feet

Given---> Distance at which Nile can throw a javelin is inversely proportional to the weight of that javelin and Nile can throw a 1.5 pound javelin exactly 260 feet.

To find ---> How far Nile can throw a two pound javelin in feet

Solution---> Let mass of javelin be m and distance of throw of javelin be d.

ATQ , d ∝ 1 / m

=> d = k ( 1 / m )

=> d = k / m

Now Let masses and distances travelled by javelin in first and second throw m₁ , m₂ and d₁, d₂ respectively then

d₁ = k / m₁ ...........................(1)

d₂ = k / m₂ ............................(2)

Dividing equation ( 1 ) and ( 2 )

d₁ / d₂ = ( k / m₁ ) ( m₂ / k )

=> d₁ / d₂ = m₂ / m₁

ATQ, d₁ = 260 feet , m₁ = 1.5 pound ,

m₂ = 2 pound

Now putting values of d₁ , m₁ and m₂ we get,

=> 260 / d₂ = 2 / 1.5

=> 2 d₂ = 260 × 1.5

=> d₂ = 260 × 1.5 / 2

=> d₂ = 130 × 1.5

=> d₂ = 195 feet

So 2.0 pound javelin travels 195 feet distance .

Answer 2

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195 feet

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