# Help on Isaac Physics Kinematics lvl 4: The Basketball Player (Part D)

I've got the answers to part A, B and C, but for D, I am totally lost! Everything I try to find the minimum speed is wrong, with no hints from isaac physics guiding me at all. Either I am close, or made some pretty massive mistakes...

Any help would be great, thanks.
(edited 3 years ago)
Original post by Scoots03
I've got the answers to part A, B and C, but for D, I am totally lost! Everything I try to find the minimum speed is wrong, with no hints from isaac physics guiding me at all. Either I am close, or made some pretty massive mistakes...

Any help would be great, thanks.

It would better to say what you have tried instead of just stating you have tried everything and they are wrong.
We would not know how to help if you did not show us your working and explain working in a clear manner.

Spoiler

Original post by Eimmanuel
It would better to say what you have tried instead of just stating you have tried everything and they are wrong.
We would not know how to help if you did not show us your working and explain working in a clear manner.

Spoiler

So In part B I'm pretty sure I worked out the range formula R=(u^2 * sin(2θ))/g ,and in part C I worked out sin(θ)=1/2. So as the max angle is 45 degrees, the minimum would be the inverse of sin=1/2 so 30 degrees, right? Subbing that in gets (u^2 * sin(60))/g = R and to get L and a into the equation I'm pretty sure it is L-a = R, as if considering the centre of the ball to be where to measure from, the shallowest angle (30) would be closer to the back of the hoop than to the front when going in, so the centre would be the distance 'a' from the back of the hoop, making the total distance L-a (this is the part I was most unsure about, it might be just L or L-3a, so I was a bit confused over this). Subbing this in gets L-a = (u^2 * sin(60))/g.

Rearranging gets g(L-a)/sin(60) = u^2. sin(60) = √3 /2 , so subbing that in gets 2g(L-a)/√3 = u^2 and finally rooting that all would be the apparent answer (for me anyway), but it's not. This is why I'm a bit confused, as even using just L or L-3a or L-2a still doesn't get the right answer.

Here's my working out on paper:
(edited 3 years ago)
Original post by Scoots03

So In part B I'm pretty sure I worked out the range formula R=(u^2 * sin(2θ))/g ,and in part C I worked out sin(θ)=1/2. So as the max angle is 45 degrees, the minimum would be the inverse of sin=1/2 so 30 degrees, right? Subbing that in gets (u^2 * sin(60))/g = R and to get L and a into the equation I'm pretty sure it is L-a = R, as if considering the centre of the ball to be where to measure from, the shallowest angle (30) would be closer to the back of the hoop than to the front when going in, so the centre would be the distance 'a' from the back of the hoop, making the total distance L-a (this is the part I was most unsure about, it might be just L or L-3a, so I was a bit confused over this). Subbing this in gets L-a = (u^2 * sin(60))/g.

Rearranging gets g(L-a)/sin(60) = u^2. sin(60) = √3 /2 , so subbing that in gets 2g(L-a)/√3 = u^2 and finally rooting that all would be the apparent answer (for me anyway), but it's not. This is why I'm a bit confused, as even using just L or L-3a or L-2a still doesn't get the right answer.

Here's my working out on paper:

Your issue lies that there is a mismatch of the shallowest angle and the range (L - a).
How did the shallowest angle occur? The shallowest angle will affect the range that we use.
Sorry I'm not too sure what you mean. So sin(Ø) has to equal ½, so surely it has to be 30°, but what does that have to do with the range? It probably has a simple answer but I can't see it.
Original post by Scoots03
Sorry I'm not too sure what you mean. So sin(Ø) has to equal ½, so surely it has to be 30°, but what does that have to do with the range? It probably has a simple answer but I can't see it.

How did you compute the shallowest angle?
Original post by Eimmanuel
How did you compute the shallowest angle?

Because it was the shallowest angle, I assumed it would barely skim across the front of the hoop as well as the back of the hoop, with the centre of the ball being at the centre of the hoop as it passed. Using this I drew out three circles showing this and drew perpendicular lines in order to make a right-angled triangle so that sin(θ)=a/2a, leading to it being 1/2.

Here's a rough re-sketch I did to show how I did it (attached file)
Original post by Scoots03
Because it was the shallowest angle, I assumed it would barely skim across the front of the hoop as well as the back of the hoop, with the centre of the ball being at the centre of the hoop as it passed. Using this I drew out three circles showing this and drew perpendicular lines in order to make a right-angled triangle so that sin(θ)=a/2a, leading to it being 1/2.

Here's a rough re-sketch I did to show how I did it (attached file)

First of all, you should realize that the range equation is applied for a projectile path NOT a straight-line path.

The ball navigates through the hoop at the shallowest angle based on a straight path NOT a projectile path and this is clearly shown in your diagram.

The range for the projectile motion is thus
$R \ne l-a$

The range based on your drawing can be approximated or treated by
$l - na$

where n is an integer.

Original post by Eimmanuel
First of all, you should realize that the range equation is applied for a projectile path NOT a straight-line path.

The ball navigates through the hoop at the shallowest angle based on a straight path NOT a projectile path and this is clearly shown in your diagram.

The range for the projectile motion is thus
$R \ne l-a$

The range based on your drawing can be approximated or treated by
$l - na$

where n is an integer.