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3 years ago

An arrow is shot at an angle 38 ◦ with the horizontal. It has a velocity of 46 m/s. How high will the arrow go?

Physics

1 answer:

Klio2033 [76]3 years ago

8 0

Answer:

40·919 m

Explanation:

Initial velocity of the arrow = 46 m/s

Angle at which it is thrown from horizontal = 38°

<h3>At the maximum height, the vertical component of velocity will be 0</h3>

Initial velocity in vertical direction = 46 × sin(38) = 28·32 m/s

From the formula

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Considering the formula in vertical direction and taking upward direction as positive

v = 0

u = 28·32 m/s

a = - g = - 9·8 m/s²

Let s be the maximum height

- 28·32² = - 2 × 9.8 × s

⇒ s = 40·919 m

∴ The arrow will go 40·919 m high

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Answer:

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Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

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h = Altitude = 35000 ft

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$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

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The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

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