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jek_recluse [69]
4 years ago
12

How does the range of the pumpkin change if its initial velocity is tripled (keeping the angle fixed and less than 90∘)? How doe

s the range of the pumpkin change if its initial velocity is tripled (keeping the angle fixed and less than )? The pumpkin's range is nine times as far. The pumpkin's range is eighteen times as far. The pumpkin's range is three times as far.
Physics
1 answer:
Deffense [45]4 years ago
5 0

Answer:

The pumpkin's range is nine times as far.

Explanation:

Given,

In the first case,

Let a pumpkin is thrown with initial velocity u with an angel theta above the horizontal axis.

Therefore the range of the pumpkin is,

R\ =\ \dfrac{u^2sin2\theta}{g}\,\,\,\,\,\,\,\,\,\,\,\,eqn(1)

Now in the second case

Initial velocity of the pumpkin is three times the first case,

\therefore U\ =\ 3u

Let R' be the new range of the pumpkin.

New range of the pumpkin is,

\therefore R'\ =\ \dfrac{U^2sin2\theta}{g}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn(2)

From eqn (1) and (2), we get,

\therefore R'\ =\ \dfrac{(3u)^2sin2\theta}{g}\\\Rightarrow R'\ =\ \dfrac{9u^2sin2\theta}{g}\\\Rightarrow R'\ =\ 9 R

Hence the pumpkin's range is nine times of the initial case.

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if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and
Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62

\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99

||v|| = 85

θ = 335°

\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03

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Resultant

R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15

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Magnitude of resultant = 131.15

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