TBR Physics P 31 and 32

[DarkJoker]

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I can't seem to understand TBR Physics pages 31 and 32. I have never ever seen this material before even in my physics textbook. Never have I used the quadratic formula in physics. Of all, I know it will not be on the MCAT. Is there any simple solution to this problem that won't take me over 3 minutes?

 

[Erythropoietin]

That's actually the only way to figure out time but I've done a bunch of projectile motion discretes/passages and never came across a situation where I had to use the quadratic formula.
The worst case you can do here is that take the longer route and find the final velocity and use another linear motion formula.
The key here is to realize that to find the horizontal distance (range), you need the math product of horizontal velocity and time, and that you can find time from the given y-direction values.

 

[MedPR]

I can't seem to understand TBR Physics pages 31 and 32. I have never ever seen this material before even in my physics textbook. Never have I used the quadratic formula in physics. Of all, I know it will not be on the MCAT. Is there any simple solution to this problem that won't take me over 3 minutes?

Who says the quadratic formula won't be on the MCAT? It could be.

 

[DarkJoker]

Who says the quadratic formula won't be on the MCAT? It could be.

Ok, maybe it could be. But is there any other way to solve this problem in less than 3 minutes?

 

[MedPR]

Ok, maybe it could be. But is there any other way to solve this problem in less than 3 minutes?

If you keep reading, on page 33 it explains the "Turbo Solution"

You can also probably use the range formula to calculate the range it travels up until it reaches the same height above the ground as it started. Then use h=5t^2 to find the time it takes to freefall from h=310, then use that time to calculate x, then add the two xs.

 

[Golaso]

If you want to use the quadratic formula u want to make sure you have theproper sign for each the displacement above the ground and the gravity sign.

In example 1.9 both are negative, because both variables are moving in thesame direction. Now some of you might think TBR or TPR recommend using apositive value for the direction the object is moving in. that's fine as longas you pick the negative time value you obtain from the quadratic equation.

i.e. according to example 1.9.

+310 m= (26.8m/s)t + 1/2 (+9.8m/s^2) t^2 (I changed the d and g values to bepositive)

Solving for t? t=+5.68 s or t=-11.15 Which one? -11.15 why? the mostexplanation i can provide is by comparing the two numbers AFTER taking the absolutefunction for each, you need to use a little bit of thinking and pick thegreater number. The two numbers, 11.15 second is the time for the object to bespent in the air and 5.68 is the time complimentary (the imaginary time)