If a body is projected with a velocity of $9.8 m / s$ making an angle of $45^{o}$ with the horizontal, then the range of the projectile is (Take $g = 9.8 m / s^{2}$ )

39.2 m

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9.8 m

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4.9 m

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19.6 m

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Solution

The correct option is D $9.8 m$
Given, $v = 9.8 m / s$
Range, $R = \frac{v^{2} sin 2 θ}{g} = \frac{(9.8)^{2} sin (2 \times 45^{0})}{9.8}$ $= 9.8 m$

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If a body is projected with a velocity of 9.8m/s making an angle of 45o with the horizontal, then the range of the projectile is (Take g=9.8 m/s2)

The correct option is D 9.8mGiven, v=9.8 m/sRange, R=v2sin2θg=(9.8)2sin(2×450)9.8=9.8 m

If a body is projected with a velocity of $9.8 m / s$ making an angle of $45^{o}$ with the horizontal, then the range of the projectile is (Take $g = 9.8 m / s^{2}$ )

The correct option is D $9.8 m$
Given, $v = 9.8 m / s$
Range, $R = \frac{v^{2} sin 2 θ}{g} = \frac{(9.8)^{2} sin (2 \times 45^{0})}{9.8}$ $= 9.8 m$