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luda_lava [24]
3 years ago
11

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 wi

ll shorten the range by 34.1 % . How far will the projectile travel in the horizontal direction, R ?
Physics
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

R=\frac{u^{2}Sin2\theta }{g}

R=\frac{381^{2}Sin147 }{9.8}

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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The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

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Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

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At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

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Vxo = 1.34 s

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Increase in range by using of halters is

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ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

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Base on my research, within 2 hours you have a number of atoms which remain. 
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Answer:

d = 4 d₀o

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