A body is projected into air with a velocity 3-4-5k- m-s- Taking X-Y plane as base and Z axis as vertical- the range of projectile g -10 m-s2

The correct option is B 5 m cosθ =55√(2)=1√(2) θ=45 |V|=√(9+16+25)=5√(2) Range =v^2sin 2θg =(√(50))^2sin 9010 =5010 =5 m. ...

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Standard XII

Physics

Helical Motion

A body is pro...

Question

A body is projected into air with a velocity $(3^i + 4^j + 5^k) m / s$ . Taking $X - Y$ plane as base and $Z$ axis as vertical, the range of projectile $(g = 10 m / s^{2})$

10 m

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5 m

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15 m

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20 m

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Solution

The correct option is B 5 m
$cos θ = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}}$
$θ = 45$
$| V | = \sqrt{9 + 16 + 25} = 5 \sqrt{2}$
Range $= \frac{v^{2} sin 2 θ}{g}$
$= \frac{(\sqrt{50})^{2} sin 90}{10}$
$= \frac{50}{10}$
$= 5$ m.

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A body is projected into air with a velocity (3^i+4^j+5^k)m/s. Taking X−Y plane as base and Z axis as vertical, the range of projectile (g=10m/s2)

The correct option is B 5 mcosθ=55√2=1√2θ=45|V|=√9+16+25=5√2Range=v2sin2θg=(√50)2sin9010=5010=5m.

A body is projected into air with a velocity $(3^i + 4^j + 5^k) m / s$ . Taking $X - Y$ plane as base and $Z$ axis as vertical, the range of projectile $(g = 10 m / s^{2})$

The correct option is B 5 m
$cos θ = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}}$
$θ = 45$
$| V | = \sqrt{9 + 16 + 25} = 5 \sqrt{2}$
Range $= \frac{v^{2} sin 2 θ}{g}$
$= \frac{(\sqrt{50})^{2} sin 90}{10}$
$= \frac{50}{10}$
$= 5$ m.