Angle that maximizes Range in projectile motion (TBR physics question)

[kevincgk]

Hello. Hoping someone might be able to help...

I understand that launching a projectile at 45 degrees maximizes the range of the projectile IF the starting and ending heights are the same. I'm having a little trouble understanding how this changes, however, when the starting and ending heights are different like when you launch a projectile off a cliff.

Particularly, The Berkeley Review (TBR) Physics Part I Chapter 1 (Translational Motion), Practice Passage IV, Question 27 (page 41). I read the explanation (page 55) which says "The best compromise of maximizing the angle to maximize the flight time, while simultaneously minimizing the angle to maximize the x-direction velocity, is achieved at a value of slightly less than 45 degrees." I get why a bigger angle increases flight time and a smaller angle increases x velocity (cos(theta)), but I don't get how we know that an angle less than 45 degrees is the right compromise between these competing factors.

Also, in the problem (page 41), each graph for answers A-D shows the range being the same for 0 degrees and 90 degrees. How is that possible? Isn't zero degrees straight ahead which would definitely have a greater range than 90 degrees which is shooting straight up in the air so the projectile would come right back down???


THANKS!!

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[BerkReviewTeach]

Hello. Hoping someone might be able to help...

I understand that launching a projectile at 45 degrees maximizes the range of the projectile IF the starting and ending heights are the same. I'm having a little trouble understanding how this changes, however, when the starting and ending heights are different like when you launch a projectile off a cliff.

Particularly, The Berkeley Review (TBR) Physics Part I Chapter 1 (Translational Motion), Practice Passage IV, Question 27 (page 41). I read the explanation (page 55) which says "The best compromise of maximizing the angle to maximize the flight time, while simultaneously minimizing the angle to maximize the x-direction velocity, is achieved at a value of slightly less than 45 degrees." I get why a bigger angle increases flight time and a smaller angle increases x velocity (cos(theta)), but I don't get how we know that an angle less than 45 degrees is the right compromise between these competing factors.

It might be easier to think about what the situation allows different than when the launch point and landing point are at the same height. In this case, because you are landing at a lower elevation than you start, the object will be in the air longer than the standard scenario. This means that flight time is increased. If you lower the launch angle slightly, you can convert some of the extra flight time into extra speed, which leads to a balance between flight time and speed in the x-direction.

Another way to think about this involves your instincts when you throw an object onto a roof. You instinctively make the launch angle greater than 45 degrees, because with an elevated landing point, the flight time has been reduced. A greater launch angle will give back some of the lost flight time. Throwing off of a cliff is the opposite situation.

Also, in the problem (page 41), each graph for answers A-D shows the range being the same for 0 degrees and 90 degrees. How is that possible? Isn't zero degrees straight ahead which would definitely have a greater range than 90 degrees which is shooting straight up in the air so the projectile would come right back down???
THANKS!!

You are right. In the graphs in my picture it looks like the difference in range between 0 and 90 is minimal.