EK Physics In-Class Exam 1, #7

[thestrokes14]

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This question is on p.161 of Physics Examkrackers In-Class Exam.

I simply cannot understand this problem (the one about why Tom must throw a ball DIRECTLY at Jim in order for him to catch it). I read the solution, and I still can't understand it. If anyone has the EK Physics book and has time to look at this question, could you explain your approach? Thanks.

 

[ljc]

(I didn't check the book, but I'm assuming it's the problem where he jumps off the diving board and catches the ball mid air)
This one took a LONG time for me to accept as well. Just draw a picture and it should become clear.

To make it easy, pick a set horizontal distance and time, and then vary the angle/vertical speed.
Let's do 10m horizontally, with 1 second of travel time. So draw a triangle with a horizontal component of 10m/s.
Now, Jim will fall for 1s before him and the ball cross planes (because we picked 1s to be our travel time), so we need to figure out Jim's falling distance in 1s. Kinematics shows us this will be 5m. So the ball must be at whatever Jim's starting height is, minus 5 meters.
Now if we know the ball is in the air for 1 second, what will its average vertical velocity change by? Well, we can pick any numbers and check this: say the ball starts at 100m/s, after 1 second it will be at 90m/s, so average these two giving 95m/s average velocity. If it starts at 40m/s, it will be at 30m/s after one second, giving 35m/s. So we can see that the ball will lose exactly 5m/s of average vertical speed in 1 second.
So since Jim falls 5 meters in 1 second, if Jim starts at 100 meters, we need the ball to meet him at 95 meters height in 1 second. Therefore, the ball must have an average velocity of 95m/s vertically.
So if we know we need an average velocity of 95s, and our travel time is 1s, what should our starting speed be? Well we know vertical velocity changes by 10m/s every second, and we're in the air for 1 second, so we need to have 100m/s and 90m/s be our starting and ending times, respectively. Therefore, start speed is 100m/s.
So now add your vertical component to the triangle, giving you a vertical side of 100m/s and a horizontal of 10m/s. If we remove the "per seconds" on the triangle, we have a 10m horizontal by 100m vertical triangle, which is exactly where Jim is standing on the board. Thus, we can see the ball must be thrown directly at Jim.

Plug in numbers and try it for yourself. But only do this once, because it is WAY too time consuming to try and prove to yourself in the middle of an exam. Once you verify it is true, just remember this:
Object A and B start from rest. Object A falls from height X, Object B is thrown and must intersect. Object B should be aimed directly at Object A's starting spot, regardless of horizontal or vertical velocity

Edit: Formatted it to make it a little more clear

 

[thestrokes14]

(I didn't check the book, but I'm assuming it's the problem where he jumps off the diving board and catches the ball mid air)
This one took a LONG time for me to accept as well. Just draw a picture and it should become clear.

To make it easy, pick a set horizontal distance and time, and then vary the angle/vertical speed.
Let's do 10m horizontally, with 1 second of travel time. So draw a triangle with a horizontal component of 10m/s. Now, Jim will fall for 1s before him and the ball cross planes (because we picked 1s to be our travel time), so we need to figure out Jim's falling distance in 1s. Kinematics shows us this will be 5m. So the ball must be at whatever Jim's starting height is, minus 5 meters. Now if we know the ball is in the air for 1 second, what will its average vertical velocity change by? Well, we can pick any numbers and check this: say the ball starts at 100m/s, after 1 second it will be at 90m/s, so average these two giving 95m/s average velocity. If it starts at 40m/s, it will be at 30m/s after one second, giving 35m/s. So we can see that the ball will lose exactly 5m/s of average vertical speed in 1 second. So if Jim starts at 100 meters, we need the ball to be at 95 meters height in 1 second. Therefore, it must have an average velocity of 95m/s vertically. This is because we figured out Jim will fall 5 meters in 1 second (which again, we picked arbitrarily - it could have been any time). So if we know we need an average velocity of 95s, and our travel time is 1s, what should our starting speed be? Well we know vertical velocity changes by 10m/s, so we need to have 100m/s and 90m/s be our starting and ending times, respectively. Therefore, start speed is 100m/s. So now add your vertical component to the triangle, giving you a vertical side of 100m/s and a horizontal of 10m/s. If we remove the "per seconds" on the triangle, we have a 10m horizontal by 100m vertical triangle, which is exactly where Jim is standing on the board. Thus, we can see the ball must be thrown directly at Jim.

Plug in numbers and try it for yourself. But only do this once. Once you verify it is true, just remember this:
Object A and B start from rest. Object A falls from height X, Object B is thrown and must intersect. Object B should be aimed directly at Object A's starting spot, regardless of horizontal or vertical velocity (assuming they start from rest)

Wow, thanks a lot! The way I initially approached it was by finding the time it took Jim to reach the pool (1.4s), and then I assumed Tom threw the ball at velocity 10m/s. I then back solved for theta and got 45 degrees...but it bugged me that I had to assume a velocity. Anyways, thanks for the concrete explanation!

 

[ljc]

Wow, thanks a lot! The way I initially approached it was by finding the time it took Jim to reach the pool (1.4s), and then I assumed Tom threw the ball at velocity 10m/s. I then back solved for theta and got 45 degrees...but it bugged me that I had to assume a velocity. Anyways, thanks for the concrete explanation!

Glad to help! Like I said, it took me forever to accept this as true. I probably did this problem with 4-5 different sets of distances/velocities before I accepted the above principle. But once you accept it and apply it, it can help you solve many MCAT style questions in 5-10s instead of 2-3mins.

 

[chiddler]

Wow, thanks a lot! The way I initially approached it was by finding the time it took Jim to reach the pool (1.4s), and then I assumed Tom threw the ball at velocity 10m/s. I then back solved for theta and got 45 degrees...but it bugged me that I had to assume a velocity. Anyways, thanks for the concrete explanation!

the velocity of 10 m/s is given in the passage!

 

[milski]

Object A and B start from rest. Object A falls from height X, Object B is thrown and must intersect. Object B should be aimed directly at Object A's starting spot, regardless of horizontal or vertical velocity

Edit: Formatted it to make it a little more clear

That! 👍 Just go with it, it will save you a lot of grief.

 

[thestrokes14]

That! 👍 Just go with it, it will save you a lot of grief.

This statement just sums up the answer. It doesn't explain why (the text above does).

 

[thestrokes14]

And actually, the point here is that the speed at which you throw the ball changes depending on WHERE you want the person diving to catch the ball. If you throw the ball at 10 m/s, as the passage suggests, then the guy diving can only catch the ball directly at ground level. If you want him to catch the ball after 1 second, as the person who responded to my post suggested, the velocity is NOT going to be 10 m/s.

 

[ljc]

And actually, the point here is that the speed at which you throw the ball changes depending on WHERE you want the person diving to catch the ball. If you throw the ball at 10 m/s, as the passage suggests, then the guy diving can only catch the ball directly at ground level. If you want him to catch the ball after 1 second, as the person who responded to my post suggested, the velocity is NOT going to be 10 m/s.

Correct. It's good to be able to think about it both ways. You're set.

 

[milski]

This statement just sums up the answer. It doesn't explain why (the text above does).

Yes, but it's an important takeaway which is generally applicable to a lot of problems. I thought it was worth calling it out for those who did not read the whole explanation.