# Physics: Projectile motion -- Maximize the distance and hang-time of a soccer ball

## Homework Statement

The question is in the pic.

## 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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nrqed
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## Homework Statement

The question is in the pic.

## 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.
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.

Yes, I think that is what I did.
X=vi*cos(angle)*t +1/2*g*t^2
Assuming some values will get me an answer of B

gneill
Mentor

## Homework Statement

The question is in the pic.

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

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

nrqed
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Yes, I think that is what I did.
X=vi*cos(angle)*t +1/2*g*t^2
Assuming some values will get me an answer of B
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?

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

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?
Yes, do u mean x=vi*cosine theta*t

gneill
Mentor
the horizontal distance = initial velocity times cosine theta x time +half times acceleration due to gravity x time squared
See what @nrqed said in post #6 above.

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

nrqed
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Yes, do u mean x=vi*cosine theta*t
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.

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

• nrqed
Note that “... certain to ...” means it has to be true for all possible choices of initial v and initial angle and for all amounts of change. There are several answers which might be true for some particular conditions, but there is only one which is true for all conditions.

• hmmm27
hmmm27
Gold Member
You're kidding, right ?

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Note that “... certain to ...” means it has to be true for all possible choices of initial v and initial angle and for all amounts of change. There are several answers which might be true for some particular conditions, but there is only one which is true for all conditions.
Yes, I think now after these two equations I can see the relation clearly. Thanks.

You're kidding, right ?
Umm?

hmmm27
Gold Member
On your picture, what do the words say underneath the diagram ? and what could that signify.

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

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

nrqed
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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.
Projectile motion refers to situations when only the force of gravity is acting. So yes, it is a projectile motion situation.

hmmm27
Gold Member
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".

• YMMMA
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
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.

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

• hmmm27 and gneill
hmmm27
Gold Member
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.

• YMMMA, nrqed and gneill
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 ?

nrqed
Homework Helper
Gold Member
In the equation, the horizontal range is directly proportional to the square of intial velocity and sine theta.
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?

• YMMMA
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..

TIme of Flight= 2v0sinθ/g and Horizontal Range=v02sin2θ/g
Looking at both the equations, it's easily concluded that by increasing v0 both Time of Flight and Horizontal Range will increase.
Not the tricky part is with θ.
For the given value of v0, Horizontal Range at θ and 90°-θ are equal although the time of flight is different. Why?
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 v0 is also constant, then Horizontal Range for θ=30° and θ=60° will be same. So it's now clear that increasing θ will not increase Horizontal Range forever.
So out of the options present, I'll put my money on Option. C

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