Two balls, A and B are thrown at the same angle above horizontal, but initial velocity of ball A is two times greater than initial velocity of ball B. Ignore air resistance. We can conclude that maximum height reached by ball A is (section 4.3) two times maximum height reached by B greater than maximum height reached by B, but we don't have enough information to give exact answer. four times the maximum height reached by B same as maximum height reached by B

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Physics

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31 August, 15:59

Two balls, A and B are thrown at the same angle above horizontal, but initial velocity of ball A is two times greater than initial velocity of ball B. Ignore air resistance. We can conclude that maximum height reached by ball A is (section 4.3) two times maximum height reached by B greater than maximum height reached by B, but we don't have enough information to give exact answer. four times the maximum height reached by B same as maximum height reached by B

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Anabelle · Answer 1

Let the initial velocity of A and B be 2U₁ and U₁ at a common angle of θ.

At maximum height H, Velocity v = 0

For vertical displacement

Vertical component = U₁ Sinθ and

v² = u² - 2gH

H = u² / 2g

H₁ = (2U₁ sinθ) ² / 2g

H₂ = (U₁ sinθ) ² / 2g

H₁ / H₂ = 4/1