For a projectile thrown into space with a speed v, the horizontal range is ((3)^(1/2)v^2)/2g. The vertical range is v^2/8g. The angle which the projectile makes with horizontal initially is (a)15° (b)30° (c)45° (d)60°

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Arun · Answer

Dear Adyasha horizontal range = u^2 sin2@ / g hence sin2@ = sqrt(3)/ 2 2@ = 60degree @ = 30 degree You can also check by height

Khimraj · Answer

max heightis given by H = v 2 sin 2 /2g = v 2 /8g so sin = ½ then = 30 o Hope it clears.............................

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For a projectile thrown into space with a speed v, the horizontal range is ((3)^(1/2)v^2)/2g. The vertical range is v^2/8g. The angle which the projectile makes with horizontal initially is (a)15° (b)30° (c)45° (d)60°

For a projectile thrown into space with a speed v, the horizontal range is ((3)^(1/2)v^2)/2g. The vertical range is v^2/8g. The angle which the projectile makes with horizontal initially is
(a)15° (b)30° (c)45° (d)60°

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For a projectile thrown into space with a speed v, the horizontal range is ((3)^(1/2)v^2)/2g. The vertical range is v^2/8g. The angle which the projectile makes with horizontal initially is (a)15° (b)30° (c)45° (d)60°

For a projectile thrown into space with a speed v, the horizontal range is ((3)^(1/2)v^2)/2g. The vertical range is v^2/8g. The angle which the projectile makes with horizontal initially is(a)15° (b)30° (c)45° (d)60°

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For a projectile thrown into space with a speed v, the horizontal range is ((3)^(1/2)v^2)/2g. The vertical range is v^2/8g. The angle which the projectile makes with horizontal initially is
(a)15° (b)30° (c)45° (d)60°