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A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction will shorten the range by 34.1 % . How far will the projectile travel in the horizontal direction, R ?

Answer :

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]

[tex]R=\frac{381^{2}Sin147 }{9.8}[/tex]

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

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