TBR Chapter 1 inconsistency?

[pfaction]

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Edit: NOPE, I CAN'T READ, IGNORE TOPIC.

Between Page 33 and Page 25. In their turbo solved 1.9, I got the answer wrong. I said that if vy at the top is 0, then if it drops 3 seconds, it's 45 meters. Nope, they said I have to use the whole y=vot + 1/2 g t(sq) method because there's an initial vy.

So on page 33, to my surprise, they allow the shortcut to be used even though there's a Vy initial.

😕😕😕 What the hell?


Also, I started this chapter at 4 PM. It's now 5:45. Nowhere near done. Holy ****.

 

[MedPR]

Between Page 33 and Page 25. In their turbo solved 1.9, I got the answer wrong. I said that if vy at the top is 0, then if it drops 3 seconds, it's 45 meters. Nope, they said I have to use the whole y=vot + 1/2 g t(sq) method because there's an initial vy.

So on page 33, to my surprise, they allow the shortcut to be used even though there's a Vy initial.

😕😕😕 What the hell?


Also, I started this chapter at 4 PM. It's now 5:45. Nowhere near done. Holy ****.

The shortcut being h=1/2gt^2? That never ever ever works if voy isn't 0. Post the two problems, maybe you are misreading something.

 

[pfaction]

Approximately how high above the ground is a spherical projectile 9 seconds after it has been launched straight up with Vy 60 m/s?

My approach: 60 m/s straight up: T = 6s at Vy=0, so my logic was 3 seconds after that; if 1s=5m and 2s=20m and 4s=80m, 3s is 45m. Nope! Have to do the whole vot+1/2at^2.

A note I'd like to add is that if it was this height at 9 seconds, it was this height at 3 seconds. So the VoY IS 0 at this point, but they say I can't use their shortcut.

Compare it to the following problem:

Next problem:
Blah blah throw an apricot off a cliff at 35 m/s angle of 50 degrees above horizontal. Cliff is 310 m tall, etc etc. The etc etc is not issue here. Cos=0.6, Sin=0.76

So here, I was like oh man I have to do that whole thing? Let me read to make sure. And it said;
For the upwards portion of the flight, we know Voy is 26.8 so the apex (Vy=0) is 2.8 sec. We can approximate that the apricot climbs about 40m to launch. Why can we approximate that? Their shortcut that didn't work for the previous problem.

And that's where I'm not okay, because I don't see why I can't use it in one problem and not for another. It's the same concept.

 

[MedPR]

Approximately how high above the ground is a spherical projectile 9 seconds after it has been launched straight up with Vy 60 m/s?

My approach: 60 m/s straight up: T = 6s at Vy=0, so my logic was 3 seconds after that; if 1s=5m and 2s=20m and 4s=80m, 3s is 45m. Nope! Have to do the whole vot+1/2at^2.

A note I'd like to add is that if it was this height at 9 seconds, it was this height at 3 seconds. So the VoY IS 0 at this point, but they say I can't use their shortcut.

Compare it to the following problem:

Next problem:
Blah blah throw an apricot off a cliff at 35 m/s angle of 50 degrees above horizontal. Cliff is 310 m tall, etc etc. The etc etc is not issue here. Cos=0.6, Sin=0.76

So here, I was like oh man I have to do that whole thing? Let me read to make sure. And it said;
For the upwards portion of the flight, we know Voy is 26.8 so the apex (Vy=0) is 2.8 sec. We can approximate that the apricot climbs about 40m to launch. Why can we approximate that? Their shortcut that didn't work for the previous problem.

And that's where I'm not okay, because I don't see why I can't use it in one problem and not for another. It's the same concept.

Yea, for the first one there is no shortcut. Just plug and chug since you know voy, time, and acceleration. Btw, the voy is only zero at one point in the first problem. It is only 0 at the apex. You can find the apex by vy^2=voy^2 - 2gy. 60^2=20y, y=180m.

What are you trying to find in the second problem? Nothing else is given? The second shortcut isn't the same as the one you are trying to apply to the first because in the first problem you aren't going from the apex downward.

time to apex = voy/2g = 28/20 = 2.8. You get 40 by using y=voyt + 1/2gt^2. Then from there it free falls so 310+40=1/2(10)t^2 because at the apex voy = 0.

 

[pfaction]

The thing I'm trying to find in the 2nd problem was the distance but that is not the point. They have to be wrong in that first question. Think about it logically.

You shoot a cannonball in a vacuum and it hits V0. Three seconds later there is 0 way it will be 130 m from the V0 point. y=halfgt2 from this point.

Example 1.17:
What are the range and maximum height for a spherical projectile launched at a 30 degree angle with respect to the horizontal with an initial of 20 m/s?
a) 32, 8
b) 45, 6
c) 26, 13
d) 34, 5
I solved this in 20 seconds.
Voy: 20m/s * (.5) = 10 m/s; therefore apex is 1 second.
1 second: 5 meters.
Boom, D, correct answer without even doing the range.

And the one before that:
Max height from 30 m/s 45 degree angle?
A) 22
B) 40
C) 55
D) 91

30(.7) is 21, 21 = 2.1 second apex, roughly 20 meters.

I don't understand why 45 isn't the answer for #1. I used their method for three other same based problems.

 

[MedPR]

The thing I'm trying to find in the 2nd problem was the distance but that is not the point. They have to be wrong in that first question. Think about it logically.

You shoot a cannonball in a vacuum and it hits V0. Three seconds later there is 0 way it will be 130 m from the V0 point. y=halfgt2 from this point.

Example 1.17:
What are the range and maximum height for a spherical projectile launched at a 30 degree angle with respect to the horizontal with an initial of 20 m/s?
a) 32, 8
b) 45, 6
c) 26, 13
d) 34, 5
I solved this in 20 seconds.
Voy: 20m/s * (.5) = 10 m/s; therefore apex is 1 second.
1 second: 5 meters.
Boom, D, correct answer without even doing the range.

And the one before that:
Max height from 30 m/s 45 degree angle?
A) 22
B) 40
C) 55
D) 91

30(.7) is 21, 21 = 2.1 second apex, roughly 20 meters.

I don't understand why 45 isn't the answer for #1. I used their method for three other same based problems.

What do you mean the cannonball hits "V0"? And why are they wrong in the first problem? You throw a ball straight up, gravity pushes it down, eventually it reaches an apex (where vy=0) then it starts to accelerate downward again.

 

[pfaction]

Right, and therefore you'd use y = vo + 1/2 gt sq.
If Vo=0
y = 1/2 g t sq
They're telling me 60 m/s is the initial launch velocity. So 6 seconds is the apex. 3 seconds after the apex is basically like saying you're dropping something and want to know how far it went in 3 seconds.
y = 5 (3sq) = 45m?

 

[MedPR]

Right, and therefore you'd use y = vo + 1/2 gt sq.
If Vo=0
y = 1/2 g t sq
They're telling me 60 m/s is the initial launch velocity. So 6 seconds is the apex. 3 seconds after the apex is basically like saying you're dropping something and want to know how far it went in 3 seconds.
y = 5 (3sq) = 45m?

I don't understand why you think 6s is the apex..

vy^2=voy^2-2gy

At apex, vy=0, therefore 2gy=60^2 where y is the apex.

20y=60^2

y=180.

Wait nevermind. 6s is the apex but that's not really important. You still use the y=voyt-1/2gt^2 equation and get something like 150m which should be the answer.

 

[pfaction]

Literally, TBR says 6 seconds in the apex. I'm copying almost verbatim.

Turbo solution: As we've seen in the previous examples, the time...reach apex is equal to initial vertical velocity divided by 9.8 (note I used 10 like everyone else). This means it takes roughly 6 seconds to reach its highest point and 12 seconds for total flight. The height it takes to reach the 9 second mark is equal to the height it reaches at 3 second mark.

6 apex; Vy = 0.
9 seconds: 3 second difference, like you're dropping something from rest and it falls for 3 seconds.

And the answer also isn't 180; I did it your way as well after I saw I got it wrong. I'm seriously WTFing here because in other problems they're not using initial velocity at all for height, like the apricot problem.

 

[milski]

6 sec is the apex - 60 m/s at 10 m/s^2 is 6s. The 45 meters that you have are from the top, not from the bottom. It will be 5*6^2-5*3^2=180-45=135 m above the ground

 

[pfaction]

My old nemesis, reading comprehension, has caused me to once again screw up a question.

Sorry guys.