Simple Physics Problem

[Murphy Durphy]

Alright, these two physics problems are pretty low-level problems concerning projectile motion and the such, but for some reason this morning I have a complete road-block and can't seem to remember anything that has to do with solving physics problems. So if someone could go ahead and explain these two to me, maybe it'll give me a kick back into gear. This is rather confusingly frusterating.

Anyway:

1) A ball is fied from the ground with an inital speed of 1.70x10(power3) m/s (about 5 times the speed of sound) at an initial angle of 55.0 degrees. Neglecting air resistance, find

a) The ball's horizontal range.
b) the amount of time the ball is in motion.


2) A daredevil is shot out of a cannon at 45.0 degrees to the horizontal with an inital speed of 25.0 m/s. A net is positioned a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

 

[sciencewhiz]

what are the 2 basic equations for projectile motion?

 

[Tylanner]

x=x

y=y

 

[Gibson486]

1)break into components...

solve b, then solve a

2) break into components.

 

[Murphy Durphy]

Originally posted by: Gibson486
1)break into components...

solve b, then solve a

2) break into components.

For 1, I have them broken into components.

vx=975
vy=1392

Problem is, I can't find a single time forumula that seems to work with calculating the distance of the bullet. It's got to incoporate gravity and initial velocity somehow, I'm just not sure what it is.

 

[RichPLS]

and drag resistance.

 

[Murphy Durphy]

And for 2, I thought just finding the vertical component would be the right thing to do, but according to the book its not.

If you draw out a triangle, shouldn't the resultant be the velocity of the daredevil (25m/s), angle is the angle he is shot out at (45 degrees) and horizontal component the distance of the net from the cannon (50m)?

Vertical component formula for y should be vy(sinangle)=25(sin45)=17.67ms.

But the book answer for this problem is 10.7m. Any ideas?

 

[Gibson486]

Originally posted by: HajikuFlip

Originally posted by: Gibson486
1)break into components...

solve b, then solve a

2) break into components.

For 1, I have them broken into components.

vx=975
vy=1392

Problem is, I can't find a single time forumula that seems to work with calculating the distance of the bullet. It's got to incoporate gravity and initial velocity somehow, I'm just not sure what it is.

Find the time it takes to hit the ground with just the y component. Use that time to solve for horizontal range range.

(hint: there are two points when v=0, when it reaches it peak height, and when it hits the ground)

 

[Murphy Durphy]

Originally posted by: Gibson486

Find the time it takes to hit the ground with just the y component. Use that time to solve for horizontal range range.

(hint: there are two points when v=0, when it reaches it peak height, and when it hits the ground)

I can't seem to figure out the formula to find time. Sigh... I feel like such a retard. I dropped out of the honors science track because I wanted a little bit easier classes this year, but here I am still stuck on relatively simple concepts.

Anyway, no time to rant. The best I can come up with is this.

To find the time it takes the bullet to hit the ground, I was using the equation t=componentx/vi,cosangle which was 975/1700cos55 which still only gives me .328 as my time when the answer is supposed to be 284 seconds.

I know thats not the right formula, but I got it from a similar problem in which a dart was fired and supposed to hit a target 10m's away, so they used 10m in place of the componentx part. I figured it would be about the same thing. But now that I think about it, 975 is probably only half of the distance it travels? Is that it's maximum height for the bullet, but not where it lands?

What exactly does the x component stand for concering a bullet that is fired?

 

[dullard]

People here hate doing homework problems, but I think I'll jump in since you are so off base.

Vertical components: Use vertical acceleration and vertical velocity here (I'm leaving off those subscripts for clarity).
Important equation: a = - g. Acceleration equals gravity since you are ignoring drag (g=9.81 m/s^2 and the negative sign means gravity pulls down).

Important equation: a = dv/dt. Acceleration is the derivative of velocity with respect to time.

Thus: dv/dt = -g. Or with a little calculus: v = v(0) - g * t . Velocity is the original velocity minus gravity multiplied by time.

Important equation: v = dy/dt. Velocity is the derivative of position with respect to time. Let y be the height above ground.

Thus: dy/dt = v(0) - g * t. Or with a little calculus: y = y(0) + v(0) * t - 1/2 * g * t^2. Note you need the initial height (data which is missing in your problem #1).

Solution to 1b: you want to know when the height is zero. Thus y = 0 meters. So solve this: 0 = y(0) + v(0) * t - 1/2 * g * t^2. All you need to know is y(0) and v(0). Finding v(0) will involve trig functions.

Hopefully that'll get you started. Yes, you should solve EVERY problem like I just did. That way, you'll be able to do this same problem from now until you die without ever having to memorize equations. Plus memorized equations like "t=componentx/vi,cosangle" often only work for one problem and not any others. Heck I don't even know for sure what that equation is trying to say.

Horizontal components: Use horizontal acceleration and horizontal velocity here (I'm leaving off those subscripts for clarity).
Important equation: a = 0. Horizontal acceleration is zero since you are ignoring drag (highly inaccurate in this case).

Important equation: a = dv/dt. Acceleration is the derivative of velocity with respect to time.

Thus: dv/dt = 0. Or with a little calculus: v = v(0). Velocity is the constant. Use trig to find the original x velocity.

Important equation: v = dx/dt. Velocity is the derivative of position with respect to time. Let x be the distance from the starting point.

Thus: dx/dt = v(0). Or with a little calculus: x = x(0) + v(0) * t. We know you start at x(0) = 0, so x = v(0) * t.

Plug in your time from problem 1b, and you are done.

 

[squeeg22]

As said before, first solve for the vertical time the projectile is in the air. The time in the air is divided into 2 equal parts - the time travelling up and the time falling back down. Neglect the horizontal component for the time being. Use the equation:

vy^2 - voy^2 = 2*a*y, where

voy^2 is the initial velocity squared; y component (1392.56 m/s)^2
vy^2 is the final vertical velocity; in this case 0 m/s since we're finding the time it takes to reach peak height
a is the acceleration constant; or 9.8 m/s^2
y is the unknown vertical distance travelled

so, we get (0 m/s)^2 - (1392.56 m/s)^2 = 2 * (-9.8 m/s^2) *y

solve for y and get 142 seconds to reach peak.
we must double this to get total time - 142s * 2 = 284 seconds.

Now, substitute this back into the general equation:

x = xo + vox*t + .5*a*t^2, solve for x

xo = initial horizontal compontent, which is zero
vox = initial horizontal velocity (cos 55 = vox/1.7e3) = 975.08m/s
a = zero, there is no acceleration force in the horizontal direction
t = time calc'd from the first part = 284 sec

you should get the simplified equation x = vox * t:

x = 975.08 m/s * 284 sec = 276922.7 m

hope this helps

ps: check my math, it's been a while

 

[squeeg22]

While I'm at it . . .

The second problem uses the same equations from the first, but in reverse order. First, you know the horizontal distance travelled.

x = xo + vox*t + .5*a*t^2

x = 50ft, the distance travelled horizontally
vox = 17.68 m/s, initial velocity compontent in the horiz. direction
a = 0 m/s^2, no horizontal accel (drag)

solve for t: 50 ft = 0 ft + 17.68 m/s*t + .5*0 m/s^2*t^2

simplifies to 50 = 17.68*t

t=2.83 sec

you want to know the final height. This is the same as problem one, except the final height is unknown, not zero (as in the peak height)

y = yo + voy*t + .5*a*t^2

yo = 0 ft, inital vertical displacement
voy = 17.68 m/s, initial velocity in the vert. directioni
t = 2.83 s, calc'd earlier
a = 9.8 m/s^2, gravity (always negative)

y = 0 ft + 17.68 m/s*2.83 sec + .5*-9.8m/s^2*(2.83 sec)^2
y = 10.71 ft, just as in your book!

 

[dullard]

Originally posted by: squeeg22
y = 10.71 ft, just as in your book!

You had a few minor typos (negative signs and mixing feet with meters), but got the mathematical answer. Now engineers step in. The body isn't a point mass and the velocity isn't constant. Thus in reality you had better use a wide net and place it BELOW 10.71 meters not right at 10.71 meters. I'd say roughly 9 meters may be a decent height for the net without any further knowledge of the cannon's accuracy.

 

[squeeg22]

Originally posted by: dullard

Originally posted by: squeeg22
y = 10.71 ft, just as in your book!

You had a few minor typos (negative signs and mixing feet with meters), but got the mathematical answer. Now engineers step in. The body isn't a point mass and the velocity isn't constant. Thus in reality you had better use a wide net and place it BELOW 10.71 meters not right at 10.71 meters. I'd say roughly 9 meters may be a decent height for the net without any further knowledge of the cannon's accuracy.

What are you talking about? In my world, all mass is a point mass. Rotational effect is always neglected. Friction is non-existent. Velocity is constant, b/c there is no such thing as drag. And, I also live in a vaccuum bubble and the answer is definitely "the moors" not the MOOPS!

 

[dullard]

Originally posted by: squeeg22
the answer is definitely "the moors" not the MOOPS!

I never liked that particular Seinfeld episode.