Physics: Projectile Motion - Maximizing Soccer Ball Distance/Hang-Time

[YMMMA]

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.

 

[nrqed]

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.

The best way is to write expressions for the Distance traveled 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.

 

[YMMMA]

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]

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.

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

 

[gneill]

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

Can you explain in words what this represents?

 

[nrqed]

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?

 

[YMMMA]

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

 

[YMMMA]

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]

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.

 

[YMMMA]

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

That’s one?

 

[nrqed]

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]

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

 

[Cutter Ketch]

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.

 

[YMMMA]

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

 

[hmmm27]

You're kidding, right ?

 

[YMMMA]

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.

 

[YMMMA]

You're kidding, right ?

Umm?

 

[hmmm27]

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

 

[YMMMA]

On your picture, what do the words say underneath the diagram ?

Do you mean ‘figure not drawn to scale’?

 

[YMMMA]

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]

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]

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".

 

[nrqed]

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]

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]

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]

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]

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..

 

[Harenderjeet Arora]

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.

 

[YMMMA]

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

You’re definitely right. I was finally able to recognize this at the end. Thank you all for your efforts...

 

[nrqed]

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

We much prefer to help students figure out the answers rather than providing them with a full solution.

By the way, I would have preferred for the OP to realize that when we plot ##\sin(2 \theta)## from 0 to 90 degrees (the range of possible angles ##\theta##), the value of ##\sin(2 \theta)## increases as ##\theta## goes from 0 to 45 degrees but then actually decreases as ##\theta## goes from 45 to 90 degrees. So as long as the angle is below 45 degrees, we gain in range by increasing it, but past 45 degrees, we start losing in range. So the optimum angle for the range is 45 degrees. This is was I was aiming at with my guiding questions.

 

[YMMMA]

We much prefer to help students figure out the answers rather than providing them with a full solution.

You certainly made me able to figure out the solution. Thanks a million.

 

[Harenderjeet Arora]

We much prefer to help students figure out the answers rather than providing them with a full solution.

I am sorry, I didn't know. Just joined the Forum now, will keep that in mind.

 

[nrqed]

I am sorry, I didn't know. Just joined the Forum now, will keep that in mind.

No problem at all! Your help will be appreciated by many students I am sure. I just wanted to give you an idea of how we work here since you just joined us. And welcome!