- #1
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 said: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...YMMMA said: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.
Can you explain in words what this represents?YMMMA said:X=vi*cos(angle)*t +1/2*g*t^2
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 said: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
the horizontal distance = initial velocity times cosine theta x time +half times acceleration due to gravity x time squaredgneill said:Can you explain in words what this represents?
Yes, do u mean x=vi*cosine theta*tnrqed said: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?
That’s one?YMMMA said: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.YMMMA said:Yes, do u mean x=vi*cosine theta*t
Yes, I think now after these two equations I can see the relation clearly. Thanks.Cutter Ketch said: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.
Umm?hmmm27 said:You're kidding, right ?
Do you mean ‘figure not drawn to scale’?hmmm27 said:On your picture, what do the words say underneath the diagram ?
It’s just to clarify that the motion of the ball is projectile.hmmm27 said:On your picture, what do the words say underneath the diagram ? and what could that signify.
Projectile motion refers to situations when only the force of gravity is acting. So yes, it is a projectile motion situation.YMMMA said: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.
It's just to clarify that the motion of the ball is projectile
YMMMA said: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.YMMMA said:So, yeah increasing both of them will certainly increase time and horizontal range based on these formulas. Answer A, then
Please, let's be respectful. People come here with various backgrounds and levels of study. We are here to help them.hmmm27 said:You're kidding, right ?
nrqed said:Please, let's be respectful. People come here with various backgrounds and levels of study. We are here to help them.
nrqed said: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.
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 said:In the equation, the horizontal range is directly proportional to the square of intial velocity and sine theta.
nrqed said: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?
If that’s right. Then, by only increasing the initial velocity, time and range would increase.YMMMA said:Doubling theta would give the same range..
Harenderjeet Arora said: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.Harenderjeet Arora said: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
nrqed said: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 said:We much prefer to help students figure out the answers rather than providing them with a full solution.
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!Harenderjeet Arora said:I am sorry, I didn't know. Just joined the Forum now, will keep that in mind.
Projectile motion is the motion of an object through the air under the influence of gravity. It is a combination of horizontal and vertical motion, and is affected by the initial velocity, launch angle, and acceleration due to gravity.
Soccer balls follow the principles of projectile motion when they are kicked. The initial velocity of the ball, along with the angle at which it is kicked, will determine the path of the ball and ultimately how far it will travel and how long it will stay in the air.
To maximize the distance and hang-time of a soccer ball, we need to optimize the launch angle and initial velocity. This can be done by finding the right balance between the two, as well as taking into account external factors such as air resistance and wind.
The ideal launch angle for a soccer ball is around 45 degrees. This angle allows for the maximum distance and hang-time, as it balances the horizontal and vertical components of the ball's motion.
The maximum distance and hang-time of a soccer ball can be calculated using the equations of projectile motion, which take into account the initial velocity, launch angle, and acceleration due to gravity. These equations can be solved using algebra or with the help of a physics calculator.