TBR Projectile Motion

[IH8ColdWeath3r]

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In TBR Physics, In passage there the answer choice for one of the problems states with a LARGER vertical velocity, the ball remains in the air LONGER (makes sense), and thus the range is greater (also makes sense bc range = Vox (t).

However, heres passage 5: A cannon located on the edge of a cliff 100 m above a flat, sandy canyon floor. Three cannon balls are launched from the cannon with the same initial speed of 50 m/s but at diff. angles. A red cannon ball is launched horizontally, a white is launched at 30 degrees angle above the horizontal, and blue at 60 degrees above the horizontal.

The blue ball has a larger max height, Because it has a LARGER VERTICAL VELOCITY. Y= Voy ^2 / 2g <==Voy for the blue ball is bigger so max height is bigger.

THe answer in the book says the white ball has a smaller maximum height (true from the analysis above) BUT A LONGER RANGE?!?! WTF 😱

someone please help.

 

[PiBond]

Since one ball is launched at 30 degrees and the other is launched at 60 degrees they will have the same range on flat land. This is because they are complement angles.

However, they are now launched from 100 meters above the ground. Thus they will still have the same range at a distance 100 meters above the ground but once they pass this region in mid-air the white ball is going to surpass the blue ball in range because it has a larger horizontal velocity.

Does that make sense? things change when we're not on flat ground

 

[BerkReviewTeach]

In TBR Physics, In passage there the answer choice for one of the problems states with a LARGER vertical velocity, the ball remains in the air LONGER (makes sense), and thus the range is greater (also makes sense bc range = Vox (t).

However, heres passage 5: A cannon located on the edge of a cliff 100 m above a flat, sandy canyon floor. Three cannon balls are launched from the cannon with the same initial speed of 50 m/s but at diff. angles. A red cannon ball is launched horizontally, a white is launched at 30 degrees angle above the horizontal, and blue at 60 degrees above the horizontal.

The blue ball has a larger max height, Because it has a LARGER VERTICAL VELOCITY. Y= Voy ^2 / 2g <==Voy for the blue ball is bigger so max height is bigger.

THe answer in the book says the white ball has a smaller maximum height (true from the analysis above) BUT A LONGER RANGE?!?! WTF 😱

someone please help.

It will help to visualize the flight in two parts: (1) from the cliff up to its highest point and then back down to the same height as the cliff (100 m above the floor) and (2) from 100 m above the floor while heading down until it strikes the ground.

For the 30-degree launch and 60-degree launch, they will both travel the same x-distance during the first part of that flight. This is because 30 degrees and 60 degrees give the same range when launched from the same elevation as which they land. Where the range difference kicks in is what happens in part (2), the 100 m descent. Both fall for the same distance (100 m), but the 60-degree launch has both a faster downward y-direction speed (meaning it will fall the 100 m in a shorter time than the 30-degree launch) and a slower x-direction speed meaning its range in that time period will be less than the 30-degree launch: R = vx(t)

 

[PiBond]

BRT was more eloquent, he wins

 

[BerkReviewTeach]

BRT was more eloquent, he wins

But you got the faster response award and explained it just as well. Kind of nice that our reasoning was on the exact same path.

Did you do that question at some point in your studies? For that question, my response was a programmed response I think, because of the picture I recall in the explanation.

 

[PiBond]

But you got the faster response award and explained it just as well. Kind of nice that our reasoning was on the exact same path.

Did you do that question at some point in your studies? For that question, my response was a programmed response I think, because of the picture I recall in the explanation.

yeah, my friends and i tried to re-create the concept by throwing rocks off the grand canyon. to each his own

 

[IH8ColdWeath3r]

Thanks pi bond and Berkteach. Both of your explanations makes things crystal clear. Do you both try to picture things as you read passages such as these, or do you try to relay the passage to a known formula. My though process on this personally relied on the equations, which was rather time consuming I must admit. Should I change my strategy and try to "visualize" what is going on....even though I am bad and this and will need practice to improve.

 

[PiBond]

Visualization is key for kinematics, at least for me. Much of MCAT physics is qualitative reasoning that can be derived much easier if you understand the situation at hand. It's extremely important to practice this during your physics mechanics review

Best of luck with your preparation 👍

 

[xtangk]

Since one ball is launched at 30 degrees and the other is launched at 60 degrees they will have the same range on flat land. This is because they are complement angles.

However, they are now launched from 100 meters above the ground. Thus they will still have the same range at a distance 100 meters above the ground but once they pass this region in mid-air the white ball is going to surpass the blue ball in range because it has a larger horizontal velocity.

Does that make sense? things change when we're not on flat ground

So, if these balls were lunched at 40 and 50 or 25 and 65, they'll still have the same range 100 meters above the ground?

 

[IH8ColdWeath3r]

So, if these balls were lunched at 40 and 50 or 25 and 65, they'll still have the same range 100 meters above the ground?

^^ Yeah great ?. I don't think this is true though because the horizontal velocity would be different because cos(40) and cos(50) are not equal. The max peak is reached at 45-degree angle

 

[PiBond]

So, if these balls were lunched at 40 and 50 or 25 and 65, they'll still have the same range 100 meters above the ground?

Yup.

let's say you launch a projectile at 40 m/s from 100 meters above the ground at angles 40 and 50. With the math below we will find that when the projectiles return to a position of 100 meters above the ground some distance away they will be at same range at that point in mid-air.

40sin(40) = 25.7...velocity in y direction
40cos(40) = 30.6...velocity in x direction

40sin(50) = 30.6...velocity in y direction
40cos(50)25.7....velocity in x direction

At the 40 degree angle we have a total time in the air of 5.14 seconds before it is 100 meters above the ground

At the 50 degree angle we have a total time in the air of 6.12 seconds before it is 100 meters above the ground

If we then solve for range (distance = x-velocity * time) and we find they are the same for both angles (because they are complementary).

of course, this relationship won't hold true once we dip below 100 meters in our example. it only works when we're dealing with "flat" land situations.