physics homework help!

[ktimekiller]

I missed a day of class, and I am slightly lost.

Problem says.

A bullet with initial velocity of 300 m/s is shot up at a 39 degrees angle.

I solved the part of the problem where it asks questions about the bullet landing in a same level ground, but I am completely lost in this next part of the question.

If the bullet landed on a platform 200 m high, how long was the bullet in the air?

I used the equation of

Vy squared = Vyo squared + 2gravity (Y-Yo)

In that case, I get the equation of

Vy = (35721 - 3920) squarerooted

Which gives me the answer of Vy = 178.

Then i plug 178 into Vy = Vyo + gravity time

Which gives me a total time of only 1 second.

Help?

 

[xcellerate]

at first glance it sounds like you're solving for how long it takes for the bullet to reach a height of 200m, not how long it takes for the bullet to peak, then fall to 200m.

 

[ktimekiller]

Yea, im trying to solve for the time of the bullet landing from a Y of 0 to a platform of 200

 

[ktimekiller]

hm, I retried with a fresh sheet of paper to make sure no mathematical errors were made, and I got a fresh answer of 37.5

Anyone getting a similar answer? if anyone is bothering?

 

[xcellerate]

okay, jeez it's been a long time since physics so correct me if I'm wrong. 300m/s at 39* is
x- 233.14m/s
y-188.79m/s

Now at the top of the path it's going 0m/s in the y-direction so we can first solve for that part of the solution with 19.26sec and that's at a height of 3542.5m. And since you aren't calculating air resistance the ball spends an equal amount of time going up as it does coming down...so it's basically 19.26*2 and then minus a little bit since it's 'stopping' at 200m. Which seems to agree with what you got...

 

[ktimekiller]

While figuring out this stuff, I got tons of "seems to agree with what I got" type of answers, but I am not 100% certain if i am right

 

[soulslicer]

200 = (300sin 39*(TIME)) + (0.5)(9.81)((TIME)^2)

Find time

simple. physics is easy.

 

[mindless_moder]

200 = (300sin 39*(TIME)) + (0.5)(9.81)((TIME)^2)

Find time

simple. physics is easy.

Vinitial is 0 at the peak so it becomes

(0.5)(9.81)((TIME)^2) = displacement

which is
time = sqrt((2*displacement)/gravity)

and Total time = time to rise + time to fall;

but i'm not sure if the displacement should be 200 or (peak height - 200)

 

[ktimekiller]

200 = (300sin 39*(TIME)) + (0.5)(9.81)((TIME)^2)

Find time

simple. physics is easy.

Well turns out my above answer was correct.

But in your equation, your left with T and Tsquared. I dont see how you solve that.

 

[vanillacrazycake]

Well turns out my above answer was correct.

But in your equation, your left with T and Tsquared. I dont see how you solve that.

The equation's slightly wrong to start off with, the acceleration due to gravity should be negative.

Using s= ut + (1/2) * a * t^2 :

200 = [(300sin39)*t] + (0.5)(-9.81)(t^2)
200 = [188.796 * t] + (-4.905) (t^2)
-4.905 t^2 + 188.796t - 200 = 0

This leaves you with a quadratic equation (ax^2 + bx + c = 0), which you can solve by using the quadratic formula: x = [ -b (plus/minus) sqrt(b^2 - 4ac) ] / 2a.
(If you currently don't know/remember how to solve quadratic equations, you should stick with the lond-winded approach of solving the equation in parts, and calculating the up and down times separately, then adding them, or better yet, learn how to do quadratic equations.)

The quadratic formula will give you two solutions: t = 1.09 or t = 37.44.
This shows that the bullet is at a height of 200m on its way up at 1.09secs after projection, and again at 37.44secs, as it comes down.

For the purposes of the question, the first value is not required, so the answer to the question is 37.44secs.

 

[ktimekiller]

ah completely ignored the fact that you could possibly use quadratics in this equation.

Thanks

EDIT: on second thought, thats one hell of a drawn out quadratics equation, i think it might be faster to solve for a single time itself using a different equation.

Unless you know a simple way of solving a quadratic equation of such type

 

[vanillacrazycake]

ah completely ignored the fact that you could possibly use quadratics in this equation.

Thanks

EDIT: on second thought, thats one hell of a drawn out quadratics equation, i think it might be faster to solve for a single time itself using a different equation.

Unless you know a simple way of solving a quadratic equation of such type

There is no other equation that can solve the question in a single go, as far as I know.

Using the quadratic formula is the single best method of solving any quadratic equation, especially one as messy and decimal place-ridden as the last one, where factorisation won't work.

Yes, the formula is a bit long and complicated to remember, but it can solve all sorts of quadratic equations, in contrast to factorisation which is only good when very specific set of numbers appear in the quadratic equation.

Not to say you shouldn't practise factorisation, as most quadratic equation specific questions you find in Maths exams are geared towards you to be able to use factorisation and find the answer nice and quick, but in a question such as the last one, where you need to use quadratics and see lots of decimal places, the automatic choice should be to use the quadratic formula. It never fails. Realize this, and hardwire it to you brain as quickly as possible.

 

[ktimekiller]

well I didnt mean solving it in a single go, but I find it quicker. Maybe I dont know a better way to solve quadratics, but if I tried to solve that, it would take me far too long. Can you show me which method of solving quadratics you would use to solve that?

As for what I find better, I would use.

Vy^2 = Vyo^2 + 2g(y-yo)

Which gets me the answer to Vy

Then I use Vy = Vyo +gt

Which gets me time.

Dont get me wrong, your method obviously works, but I have no clue how you would effectively use quadratics in such long drawn out numbers, im totally lost.

 

[TheHamster22]

Ok so you go:

Find u = 188.89m/s

Find the time it takes to peak is 19.25s, using v = u + at

The find the velocity of the bullet when it drops from its peak to 200 metres, using v^2 = u^2 + 2as. (and s = ut + 0.5at^2 ). v=178.14m/s

Use v = u + at to find the time it takes the bullet to drop 200 metres. t = 18.17s.

Therefore the total trip time of the bullet it is: T = 37.41s.

Which you almost got right, apart from being 0.1s out, gawd! >_> <_<

 

[Sandspider]

The equation's slightly wrong to start off with, the acceleration due to gravity should be negative.

Using s= ut + (1/2) * a * t^2 :

200 = [(300sin39)*t] + (0.5)(-9.81)(t^2)
200 = [188.796 * t] + (-4.905) (t^2)
-4.905 t^2 + 188.796t - 200 = 0

This leaves you with a quadratic equation (ax^2 + bx + c = 0), which you can solve by using the quadratic formula: x = [ -b (plus/minus) sqrt(b^2 - 4ac) ] / 2a.
(If you currently don't know/remember how to solve quadratic equations, you should stick with the lond-winded approach of solving the equation in parts, and calculating the up and down times separately, then adding them, or better yet, learn how to do quadratic equations.)

The quadratic formula will give you two solutions: t = 1.09 or t = 37.44.
This shows that the bullet is at a height of 200m on its way up at 1.09secs after projection, and again at 37.44secs, as it comes down.

For the purposes of the question, the first value is not required, so the answer to the question is 37.44secs.

I got this also... exact same day!

 

[Kula Meenur]

Wasn't there a dedicated thread for this???