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The horizontal range of a projectile projected with a velocity V at angle 15° with the horizontal is 50 m. What will be its horizontal range if its angle of throw is increased by 30° keeping the velocity of throw to be the same ? |
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Answer» 60 m or `R=u^(2)/g sin2theta=u^(2)/gxxsin30^@` `50=u^(2)/gxx1/2` `:. U^(2)/g=100m` When the ANGLE is INCREASED by `30^@` i.e. 30° +15° = 45° Then `R=u^(2)/gsin90^@=u^(2)/g` `:.R=u^(@)/g=100m` |
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