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pychu [463]
3 years ago
10

You throw a snowball horizontally. It hits a target 2 meters away. You notice that it hit o.9m below where you aimed. With what

horizontal velocity did you throw the snowball?
Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

The snowball travels a horizontal distance <em>x</em> and has height <em>y</em> at time <em>t</em> according to

<em>x</em> = <em>vt</em>

<em>y</em> = 0.9 m - 1/2 <em>gt</em> ²

where <em>g</em> = 9.8 m/s² is the magnitude of the acceleration due to gravity.

Set <em>y</em> = 0 and solve for <em>t</em> :

0 = 0.9 m - 1/2 <em>gt</em> ²

<em>t</em> ² = (1.8 m) / <em>g</em>

<em>t</em> ≈ 0.43 s

The target is 2 m away from where the snowball is thrown, so it was thrown with speed <em>v</em> such that

2 m = <em>v </em>(0.429 s)

<em>v</em> = (2 m) / (0.43 s)

<em>v</em> ≈ 4.7 m/s

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(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

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(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

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v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

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