TBR Projectile "turbo method"

[lDanny]

---

Members do not see this ad.

I tried to figure out how they did the solution but I get a confused how they got some of the numbers.

For example 1.14, how did they get that 1s = 5m and so on? And 45 degrees is 4 times max height?

For example 1.15, how did they approx the height?

Thanks...

 

[PiBond]

I tried to figure out how they did the solution but I get a confused how they got some of the numbers.

For example 1.14, how did they get that 1s = 5m and so on? And 45 degrees is 4 times max height?

For example 1.15, how did they approx the height?

Thanks...

It's a shortcut. The range of a projectile fired at 45 degrees is 4 times its maximum height (a close approximation). TBR uses a convention to show that the maximum possible height (h) is obtain by launching a projectile at 90 degrees (straight up) while the maximum possible range of 2h is achieved by launching a projectile at 45 degrees. We discover that at 45 degrees, the maximum height is h/2. Thus, 4 times (h/2) is 2h which is it's range.

Example 1.14 looks a little confusing in my opinion. They got the 1s = 5m from the distance covered in 1 second when you drop an object (approximately). Here's how I would solve it using TBR's turbo convention.
delta👍 = 1/2at^2
We know that the projectile is in flight for 3 seconds so it must hit its apex at 1.5. This will serve as our time.
delta👍 = 1/2(10)(1.5^2)
We find that its y-displacement is about 11.1ish
We know using TBR logic that range is 4 times max height. Therefore we get 44m for our answer (choice B).

1.15: The approximated by the height using the kinematic V = Vo + at. Our y velocity is 26.8 and at the top of its path it is zero. Therefore, 26.9/9.8 (acceleration due to gravity) gives us the time needed to reach its apex. They say its apex is 40m (using this idea of "climb time"...wtf?) but I got around 36m. Once again, they're just showing how to use a shortcut.

 

[lDanny]

Thanks for replying. Ya, their shortcut seems to be a little confusing. How did they approx the distance drop with the information provided? (1s= 5m, 2s= 20m) Thanks again for the help👍

 

[godcyning]

The 1s = 5m, 2s = 20m, ... is a general approximation for any object dropped with no initial velocity. It comes from

a = -9.8
v = -9.8t + initial velocity, which is zero
h = -4.9t^2, call it -5t^2.

So after 1 second, h = -5, 2s gives -20m, and so forth.

 

[lDanny]

ahhhh... (ah-ha moment) now i feel stupid for asking.. lol.. 😀