TBR Physics Kinematics Ex 1.14

[MrNeuro]

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Q: What is the range for a projectile launched at 45 deg angle that has a flight time of 3 seconds?

A. 14.2m
B. 44.1m
C. 72.2m
D. 144.4m

the berkley review "turbo-solution" is
With a flight time of 3 seconds, it takes 1.5 seconds to reach its apex. A 1 second drop time covers 5m and a 2-second drop time covers 20m, so a 1.5 second drop covers about 11 to 12m, so the max height is about 11-12m. The range for a 45 deg launch is four times the max height so R ~ 44-48 m.

my main issue begins at

"a 1.5 second drop covers about 11 to 12m"

so you know that in 1 second an object will fall 5 m and in 2 seconds it will fall 20 m and so intuitively you say that 1.5 is halfway between 1 and 2 thus the halfway point between 5 and 20 is 12.5. Thus, @ 1.5 seconds an object has fallen 12.5 meters. However this is different than there approximation of 11 to 12 meters. I know that if you actually solver for the distance using s=1/2at^2 that you'll find that s = 11.21, but this would take more time and hte point of this question is to solve it in a short amount of time. After this im fine i understand that @ 45 degrees the ratio of R/h is 4 and that R=44-48 if you use 11 and 12 as your estimation of h, however if you use 12.5 you find that R=50. Hence you can see my frustration that on some questions this discrepency may be the difference between a right and wrong question.

Please help,
Sincerely disgruntled mcat struggler 😡

 

[Pre-Med Village]

You can save a lot of time with these kinds of back-of-the-envelope calculations. However, to use these kinds of strategies successfully you need to have an intuitive feeling for the range of error propagating through the calculations and how that compares to the spacing of the answers. In this case, the wide spacing of the possible answers gives you permission to be more approximate. If the answers were closely spaced, then you would need to use the kinematical equations.

 

[BerkReviewTeach]

the berkley review "turbo-solution" is
With a flight time of 3 seconds, it takes 1.5 seconds to reach its apex. A 1 second drop time covers 5m and a 2-second drop time covers 20m, so a 1.5 second drop covers about 11 to 12m, ...

...and so intuitively you say that 1.5 is halfway between 1 and 2 thus the halfway point between 5 and 20 is 12.5. Thus, @ 1.5 seconds an object has fallen 12.5 meters.

As you pointed out in your solution, the distance depends on 0.5at^2. The relationship of distance to time squared is why you can't simply take the median between times to be the median between distances. If the object is continually speeding up, then it will travel more distance from the 1.5-s mark to the 2.0-s mark than it travels from the 1.0-s mark to the 1.5-s mark. You know that 12.5 would be the linear mean and that it covers less distance in the first half, so guessing a value between 11 and 12 is a good extimate that will narrow it down to a best choice.

And as Mr. Wetzel pointed out, the answer choice proximity will dictate just how much you can estimate when doing a calculation. Even with your answer of R = 50, there is a wide enough split between 44.1 and 72.2 that you can make a good choice.

You can save a lot of time with these kinds of back-of-the-envelope calculations. However, to use these kinds of strategies successfully you need to have an intuitive feeling for the range of error propagating through the calculations and how that compares to the spacing of the answers. In this case, the wide spacing of the possible answers gives you permission to be more approximate. If the answers were closely spaced, then you would need to use the kinematical equations.

Great points. That's pretty much the way BR words it in their explanations. Tricks are great for speed, but you need to use some mathematical common sense when applying them. It's not a bad idea to start with the math in your head approach to get an answer quickly and then look at the answer choices to see if there is a wide enough range between your best choice and the next closest choices. If it works then great. If not, well you spent 10 seconds to get in the ballpark which should help you when you do the longer version of the calculation to not make a careless error.

 

[ljc]

Correct me if I'm mistaken, but isn't there a much easier way to do this?

Flight time of 3s = 1.5 up, 1.5 down
1.5 up means 15m/s initial vertical velocity
Because the angle is 45 degrees, we know horizontal velocity and vertical velocity are equal. Thus, horizontal velocity is 15m/s. 15m/s * 3 seconds total flight time = 45meters range.
44.1 is the closest to this, so they must either not be using 10m/s^2 for g, or else they are including wind resistance. Either of those will lower the answer from 45 slightly, thus getting you around the stated 44.1.

This is the method I've intuitively developed from running through EK 100 times. Maybe their method is different from TBR?

 

[Hexon]

Correct me if I'm mistaken, but isn't there a much easier way to do this?

Flight time of 3s = 1.5 up, 1.5 down
1.5 up means 15m/s initial vertical velocity
Because the angle is 45 degrees, we know horizontal velocity and vertical velocity are equal. Thus, horizontal velocity is 15m/s. 15m/s * 3 seconds total flight time = 45meters range.
44.1 is the closest to this, so they must either not be using 10m/s^2 for g, or else they are including wind resistance. Either of those will lower the answer from 45 slightly, thus getting you around the stated 44.1.

This is the method I've intuitively developed from running through EK 100 times. Maybe their method is different from TBR?

a good method, but can u explain to me how "1.5 up means 15m/s initial vertical velocity"?

 

[MT Headed]

a good method, but can u explain to me how "1.5 up means 15m/s initial vertical velocity"?

v = a x t = 10 x 1.5 = 15

ljc is obviously a distinguished graduate of EK physics. EK audio osmosis will teach you to do this kind of stuff in your head.

 

[Morsetlis]

Correct me if I'm mistaken, but isn't there a much easier way to do this?

Flight time of 3s = 1.5 up, 1.5 down
1.5 up means 15m/s initial vertical velocity
Because the angle is 45 degrees, we know horizontal velocity and vertical velocity are equal. Thus, horizontal velocity is 15m/s. 15m/s * 3 seconds total flight time = 45meters range.
44.1 is the closest to this, so they must either not be using 10m/s^2 for g, or else they are including wind resistance. Either of those will lower the answer from 45 slightly, thus getting you around the stated 44.1.

This is the method I've intuitively developed from running through EK 100 times. Maybe their method is different from TBR?

A genius method. If you could not have quickly derived this, you could have also used the general kinematics equation

[/IMG]

This would've taken a bit longer since you'd have to juggle around equations and identify which ones match your variables, but would've gotten you to the same result.

 

[Chrome19]

The actual answer is actually 44.1 if g = 9.8

My approach

V=Vo+(-g)t

At max height, V=0 m/s so that t = Vo/g. Note that this t =1.5s (half of the journey)

In the y direction Vo = Vo.sin45, so t =Vo.sin45/g is the time for half the flight (1.5s).

Range (in x direction) = Vo(x).2t = Vo.cos45*2t = Vo.cos45*2*1.5s

but

Vo.sin45=Vo.cos45 = gt

therefore

Range = gt*2t = 10*1.5*2*1.5 = 20*(3/2)*(3/2) = 45 m/s

 

[BerkReviewTeach]

Correct me if I'm mistaken, but isn't there a much easier way to do this?

Flight time of 3s = 1.5 up, 1.5 down
1.5 up means 15m/s initial vertical velocity
Because the angle is 45 degrees, we know horizontal velocity and vertical velocity are equal. Thus, horizontal velocity is 15m/s. 15m/s * 3 seconds total flight time = 45meters range.
44.1 is the closest to this, so they must either not be using 10m/s^2 for g, or else they are including wind resistance. Either of those will lower the answer from 45 slightly, thus getting you around the stated 44.1.

The method you suggest is presented in other example questions in BR, because the basic philosophy is to present multiple methods for solving similar problems on the notion that different people find different methods to their liking. BR usually presents a set of similar questions that present solutions using the tradional method, a shortcut math method, and a visualization method for the various types of problems.

The shortcut math method you suggest has the following steps:

• I. Cut the 3s flight time into an up half and down half of 1.5 s each

• II. From 1.5 s of flight to reach the apex, the inital y-direction velocity must have been 15 m/s (using the shortcut kinematics equation that v = 10t)

• III. With a launch angle of 45 degrees, you know vinit-y = vinit-x

• IV. Traveling with a vx of 15 m/s for 3 seconds leads to 45 m for the range.

This is a nice 4-step method that works well and is intuitive once you practice it a few times. It can be seen in various BR questions such as #18 of the homework.

The visualization method has the following steps:

• I. Cut the 3s flight time into an up half and down half of 1.5 s each

• II. From 1.5 s of flight to reach the apex, the maximum height is about 11 m (from estimation using hmax = 5texp2 to get 1s : 5m and 2s : 20m)

• III. For 45-degree laucnhes, R = 4 x hmax, so R = 44

This is a nice 3-step method that works really well too. It shares the same first step as the shortcut math method. In my opinion, it is easier than the shortcut method you suggest because it is fewer steps and slightly easier math (thus I would answer "no" to your initial question.) But people should study all possible methods and find the one that fits their liking. That's the primary reason BR presents so many different ways to solve problems. Once you find your method, which you have, then use it repeatedly to hone it in (which you've done quite nicely).

 

[UrshumMurshum]

Correct me if I'm mistaken, but isn't there a much easier way to do this?

Flight time of 3s = 1.5 up, 1.5 down
1.5 up means 15m/s initial vertical velocity
Because the angle is 45 degrees, we know horizontal velocity and vertical velocity are equal. Thus, horizontal velocity is 15m/s. 15m/s * 3 seconds total flight time = 45meters range.
44.1 is the closest to this, so they must either not be using 10m/s^2 for g, or else they are including wind resistance. Either of those will lower the answer from 45 slightly, thus getting you around the stated 44.1.

This is the method I've intuitively developed from running through EK 100 times. Maybe their method is different from TBR?

This is what I initially thought of solving this problem. Is it concerning that I was freaking out that I wasn't seeing 45 m/s as an answer choice? I guess I've been doing rounding so much I forgot acceleration due to gravity isn't exactly 10m/s^2.

I guess the way I can explain it is this.

First the 45 degree triangle. the ratio of the vertical:horizontal:hypotenuse in a 45 degree triangle is 1:1:Sqrt(2) This tells you that whatever your vertical component is, must also be your horizontal component for any 45 degree triangle.

We can make our triangle represent velocities.

Under the influence of gravity, it takes one second to decrease vertical velocity 10 m/s. So if you had an upward velocity of 100 m/s it would take 10 seconds to reduce that 100 to 0 m/s. Think of it as decrements. Each second you're "decrementing" the m/s down 10.

Now in this problem you're told the total flight time is 3 seconds. The time they give you is the total flight time, which includes the time for the ball to go up and then back down to the ground. The time to reach the maximum height is the same amount of time for it to go back down to its initial height. (tup = tdown) (ttotal = tup + tdown) therefore (ttotal = 2tup) (tup = 1/2ttotal)

So then you can divide the total flight time in half to get the time it takes to travel to maximum height.

3/2 = 1.5

Now invert the logic I said earlier.

15 m/s 1st second goes down to 5 m/s .5 seconds goes down 5 m/s to bring you to 0 m/s which is the vertical speed it must travel when it's at is maximum height.

Now that you have the vertical component of velocity, put it on your velocity triangle to find your horizontal velocity component. Since it's a 1:1 ratio the horizontal velocity component must also be 15 m/s.

15 m/s * 3 seconds = 45 meters.

This is a little overestimated since gravity is actually slightly less than 10 m/s^2

 

[BerkReviewTeach]

Is this seriously easier than saying:

1) 1.5 is half of 3.
2) Objects fall from rest about 11 m in 1.5 seconds.
3) Range is 4 x hfallen from highest point for a 45-degree lanuch = 44.