Question

Three identical balls are thrown with the same speed at angles of 15°, 45°, and 75° with the horizontal respectively. The ratio of their distances from the point of projection to the where they hit the ground will be...

Answer 1

Explanation:

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Answer 2

Given :-

Three identical balls are thrown with same speed.

$\theta_{1} = 15°$

$\theta_{2} = 45°$

$\theta_{3} = 75°$

Hence, In this question we need to use Range of Projectile.

R = u² Sin 2Ø/g

For First Case when Ø = 15°,

${R}_{1} = \dfrac{{u}^{2} Sin 2\theta_{1}}{g}$

R1 = u² Sin 2(15°)/g

R1 = u² Sin 30°/g

R1 = u²/2g

${R}_{2} = \dfrac{{u}^{2} Sin 2\theta_{2}}{g}$

R2 = u² Sin 2(45°)/g

R2 = u²/g (Sin 90° = 1)

${R}_{3} = \dfrac{{u}^{2} Sin 2\theta_{3}}{g}$

R3 = u² Sin 2(75°)/g

R3 = u² Sin 150°/g

R3 = u² Sin (180° - 30°)/g

R3 = u²/2g

R1 : R2 : R3 = 1 : 2 : 1