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Vsevolod [243]
3 years ago
5

A projectile is fired at ????0=355.0 m/s at an angle of theta=68.4∘ , with respect to the horizontal. Assume that air friction w

ill shorten the range by 35.1% . How far will the projectile travel in the horizontal direction, ???? ?
Physics
1 answer:
Deffense [45]3 years ago
6 0

Answer:

Range will be 5707.364 m      

Explanation:

We have given initial velocity of projectile u = 355 m/sec

Angle of projection = 68.4°

Acceleration due to gravity g=9.81m/sec^2

Horizontal range is given by R=\frac{u^2sin2\Theta }{g}=\frac{355^2sin(2\times 68.4^{\circ})}{9.81}=8794.09m

It is given that range is shorten by 35.1%

So range will be 8794.09-(8794.09\times \frac{35.1}{100})=5707.364m

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7 0
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part 1

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3 years ago
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