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ella [17]
3 years ago
8

A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of th

e projectile is 1.00 m/s, and at its maximum height the speed of the projectile is 0.50 m/s. What is the angle θ ?
Physics
1 answer:
Tamiku [17]3 years ago
3 0

Answer:

the angle is about 67.79 degrees

Explanation:

We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)

We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

half\,\,max-height = \frac{v_{yi}^2}{4\,g}

we can use this information to find the y component of the velocity at that height via the formula:

v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o

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Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

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The force F₁is constant acceleration is also a constant.

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a_{1}=0.083\ m/s^2

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F_{2}=ma_{2}

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The speed of second tugboat is

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We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

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a_{3}=0.305\ m/s^2

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0.305=a_{2}+0.083

a_{2}=0.305-0.083

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Dividing force F₁ by F₂

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\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

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Explanation:

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Answer:

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Explanation:

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