Help me study for my physics test tomorrow!

midnight rain · 02-07-2012, 07:13 PM

Quote:

Originally Posted by Freebase Dali

Word. Ok. So I have the basics. Now, when doing 2 dimensional kinematics, that trig comes into play, but I also have to remember some formulas. I can put those on my phone, so I'm not worried too much about it, since my phone is also my calculator, which is allowed.

So, how would you guys go about solving something like this (taken from a homework problem):

A projectile is launched from and returns to ground level. Air resistance is absent. The horizontal range of the projectile is R = 175 m, and the horizontal component of the launch velocity is Vox = 25 m/s. Find the vertical component Voy of the launch velocity.

I haven't taken college physics but isn't the equation just displacement = velocity * time ?

Freebase Dali · 02-07-2012, 07:16 PM

Quote:

Originally Posted by Tuna

I haven't taken college physics but isn't the equation just displacement = velocity * time ?

Well displacement is actually velocity over time (d = v/t) if I remember correctly, but that's one dimensional. The problem has both the x and y dimension, and is asking for the components of the vector, which the vector would be a diagonal line. I would need to figure out the x component and the y component that made the resultant vector.

skaltezon · 02-07-2012, 07:24 PM

Quote:

Originally Posted by Freebase Dali

Well displacement is actually velocity over time (d = v/t) if I remember correctly, but that's one dimensional. The problem has both the x and y dimension, and is asking for the components of the vector, which the vector would be a diagonal line. I would need to figure out the x component and the y component that made the resultant vector.

I think Tuna's right with d = v * t .

midnight rain · 02-07-2012, 08:06 PM

Quote:

Originally Posted by Freebase Dali

Word. Ok. So I have the basics. Now, when doing 2 dimensional kinematics, that trig comes into play, but I also have to remember some formulas. I can put those on my phone, so I'm not worried too much about it, since my phone is also my calculator, which is allowed.

So, how would you guys go about solving something like this (taken from a homework problem):

A projectile is launched from and returns to ground level. Air resistance is absent. The horizontal range of the projectile is R = 175 m, and the horizontal component of the launch velocity is Vox = 25 m/s. Find the vertical component Voy of the launch velocity.

Alright man, I looked up 2 dimensional kinematics and I'm still not sure how to solve this problem. My best guess:

Using the equation: y=v*t+.5*a*t^2

where:
y = vertical displacement
v= velocity
t= time
a = acceleration

and figuring out that t=7

you have:

0 = v*7+.5*(-9.8)*7^2

which got me

v= 34.3 m/s

have no idea if that's actually right though

Freebase Dali · 02-07-2012, 08:28 PM

Quote:

Originally Posted by Tuna

Alright man, I looked up 2 dimensional kinematics and I'm still not sure how to solve this problem. My best guess:

Using the equation: y=v*t+.5*a*t^2

where:
y = vertical displacement
v= velocity
t= time
a = acceleration

and figuring out that t=7

you have:

0 = v*7+.5*(-9.8)*7^2

which got me

v= 34.3 m/s

have no idea if that's actually right though

Ah yes, that makes sense.
Seeing the equation written that way (instead of the way my professor wrote it) makes things a lot easier. It's just a matter of plug and play then.

midnight rain · 02-07-2012, 08:32 PM

Quote:

Originally Posted by Freebase Dali

Ah yes, that makes sense.
Seeing the equation written that way (instead of the way my professor wrote it) makes things a lot easier. It's just a matter of plug and play then.

Yeah which equation you use is really all dependent on what you're given in the first place.

Here's the list I got it from in case you don't have them all:

Horizontally Launched Projectile Problems

midnight rain · 02-07-2012, 07:02 PM

Ok a little heads up I noticed with the dumb google calculator

tan(20 degrees) is actually equal to 0.36, not sure why google calculator was giving me that other number. But the steps are the same. So the first answer is actually something like 27.

And yes you're answer looks right

Freebase Dali · 02-07-2012, 07:05 PM

Oh, one more thing about trig real quick. If I want to find the angle of a right triangle, I know I need to use inverse sin, cos, or tan.

Walk me through the steps of finding the angle of a right triangle whose hypoteneus is 5 units, and whose opposite side is 7 units.

midnight rain · 02-07-2012, 07:09 PM

Quote:

Originally Posted by Freebase Dali

Oh, one more thing about trig real quick. If I want to find the angle of a right triangle, I know I need to use inverse sin, cos, or tan.

Walk me through the steps of finding the angle of a right triangle whose hypoteneus is 5 units, and whose opposite side is 7 units.

I think it's as simple as arcsin(7/5). Just put that into your calculator.

It's basically just the reverse, there's no steps involved, unless it was values used on 45-45-90 triangles or 30-60-90

Plus what skaltezon said LOL

Burning Down · 02-07-2012, 07:12 PM

Quote:

Originally Posted by Freebase Dali

Oh, one more thing about trig real quick. If I want to find the angle of a right triangle, I know I need to use inverse sin, cos, or tan.

Walk me through the steps of finding the angle of a right triangle whose hypoteneus is 5 units, and whose opposite side is 7 units.

Hypotenuse is always smaller than the opposite. If I recall correctly, you can plug in arcsin(7/5) in a calculator to get your answer.

edit: damn, I was too slow

02-07-2012, 07:05 PM	#8 (permalink)
Freebase Dali Partying on the inside Join Date: Mar 2009 Posts: 5,584	Oh, one more thing about trig real quick. If I want to find the angle of a right triangle, I know I need to use inverse sin, cos, or tan. Walk me through the steps of finding the angle of a right triangle whose hypoteneus is 5 units, and whose opposite side is 7 units. __________________

02-07-2012, 07:02 PM	#7 (permalink)
midnight rain Music Addict Join Date: Mar 2009 Posts: 1,711	Ok a little heads up I noticed with the dumb google calculator tan(20 degrees) is actually equal to 0.36, not sure why google calculator was giving me that other number. But the steps are the same. So the first answer is actually something like 27. And yes you're answer looks right