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- #2

nrqed

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The best way is to write expressions for the Distance travelled as a function of ##v_0##, ##\theta_0## and g. And then to write the equation for the time spent in the air as a function of the same parameters. Then the answer will become clear.## Homework Statement

The question is in the pic.

## Homework Equations

## The Attempt at a Solution

I answered it B. Still i am not quite sure whether it is B or E. I assumed some numbers, though.

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X=vi*cos(angle)*t +1/2*g*t^2

Assuming some values will get me an answer of B

- #4

gneill

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Please elaborate on your reasoning for each choice. Helpers won't simply confirm or deny what could well just be a guess...## Homework Statement

The question is in the pic.

## Homework Equations

## The Attempt at a Solution

I answered it B. Still i am not quite sure whether it is B or E. I assumed some numbers, though.

- #5

gneill

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Can you explain in words what this represents?X=vi*cos(angle)*t +1/2*g*t^2

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nrqed

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You are mixing things along x and along y. This is not the correct equation for the X position. Have you seen the range formula?

X=vi*cos(angle)*t +1/2*g*t^2

Assuming some values will get me an answer of B

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the horizontal distance = initial velocity times cosine theta x time +half times acceleration due to gravity x time squaredCan you explain in words what this represents?

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Yes, do u mean x=vi*cosine theta*tYou are mixing things along x and along y. This is not the correct equation for the X position. Have you seen the range formula?

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That’s one?Yes, do u mean x=vi*cosine theta*t

- #11

nrqed

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Yes, this is correct. But now you want an expression in term of ##\theta_0##, ##v_0## and g only. So you don't want time. You will have to find an expression for the time of flight.Yes, do u mean x=vi*cosine theta*t

- #12

gneill

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Do google searches on "Projectile Range Formula" and "Projectile Time Of Flight Formula".

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hmmm27

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Last edited:

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Yes, I think now after these two equations I can see the relation clearly. Thanks.

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Umm?You're kidding, right ?

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hmmm27

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On your picture, what do the words say underneath the diagram ? and what could that signify.

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Do you mean ‘figure not drawn to scale’?On your picture, what do the words say underneath the diagram ?

Last edited by a moderator:

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It’s just to clarify that the motion of the ball is projectile.On your picture, what do the words say underneath the diagram ? and what could that signify.

Note: Its an SAT question. Most of the diagrams are not drawn to scale, its just for visualization.

- #21

nrqed

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Projectile motion refers to situations when only the force of gravity is acting. So yes, it is a projectile motion situation.It’s just to clarify that the motion of the ball is projectile.

Note: Its an SAT question. Most of the diagrams are not drawn to scale, its just for visualization.

- #22

hmmm27

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It's just to clarify that the motion of the ball is projectile

huh ? Sorry, I keep forgetting the amount of ESL students in here... okay, what it boils down to is you can't tell from looking at the diagram what theta is. Also, the solution choices don't give you control over the amount of increase/decrease of theta or init-velocity.

But mostly, as Cutter pointed out : "certain to".

- #23

nrqed

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So, yeah increasing both of them will certainly increase time and horizontal range based on these formulas. Answer A, then

How did you reach that conclusion? Explain carefully your reasoning and we will be able to point out where your mistake is. That's the best way to learn.So, yeah increasing both of them will certainly increase time and horizontal range based on these formulas. Answer A, then

- #24

nrqed

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Please, let's be respectful. People come here with various backgrounds and levels of study. We are here to help them.You're kidding, right ?

- #25

hmmm27

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Please, let's be respectful. People come here with various backgrounds and levels of study. We are here to help them.

You're right, of course. Post modified. Apologies to those adversely affected.

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How did you reach that conclusion? Explain carefully your reasoning and we will be able to point out where your mistake is. That's the best way to learn.

In the equation, the horizontal range is directly proportional to the square of intial velocity and sine theta.

So, by increasing the anle and initial velocity, horizontal raneg is to increase. That appears in the time flight formula,too. Am I wrong ?

- #27

nrqed

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This is not quite right. It is proportional to sin of *twice* the angle. Theta can take what values? And over that range of value, what does the function ##\sin (2 \theta)## look like?In the equation, the horizontal range is directly proportional to the square of intial velocity and sine theta.

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This is not quite right. It is proportional to sin of *twice* the angle. Theta can take what values? And over that range of value, what does the function ##\sin (2 \theta)## look like?

Doubling theta would give the same range..

- #29

Looking at both the equations, it's easily concluded that by increasing v

Not the tricky part is with θ.

For the given value of v

Because in the equation of Horizontal Range, we have sin2θ and you can check that value of sin2θ for θ=30° and θ=60° are same. Hence if v

So out of the options present, I'll put my money on Option. C

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If that’s right. Then, by only increasing the initial velocity, time and range would increase.Doubling theta would give the same range..

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