Projectile Motion (Initial Velocity)

[Milkster18]

Hi, I am trying to find the initial velocity of this question:

Using a video camera, it is observed that when a ball is kicked from A, it just clears the top of a wall at B as it reaches its maximum height. Knowing the distance from A to the wall is 20m and the wall is 4m high, determine the initial velocity at which the ball was kicked. Neglect size of the ball ( Ans Ua = 23.9m/s).

So I am thinking to find the angle you have to use trig :

so 20^2 + 4^2 = square root of 416 = 20.4m (hypotenuse)

using Sin, 4/ 20.4 = Invert Sin 0.196 = 11.3 degrees angle the ball is projected.

Now i have then angle what else do i need to find in order to find the initial velocity at which the ball was kicked?

Thanks for any help

 

[Doc Al]

So I am thinking to find the angle you have to use trig :

so 20^2 + 4^2 = square root of 416 = 20.4m (hypotenuse)

using Sin, 4/ 20.4 = Invert Sin 0.196 = 11.3 degrees angle the ball is projected.

That angle is not the angle of the ball's initial velocity. The ball travels a parabolic trajectory, not a straight line.

Analyze the vertical motion first. Figure out the vertical component of the initial velocity.

 

[Milkster18]

Ah ok, i thought about the ball not traveling in a straight line assuming the angle was wrong at some point.

Ok so if 'm trying to analyze the vertical component of the initial velocity I am going to use equation:

Sy = Uyt + 0.5(ay)t^2 Or

Vy^2 = Uy^2 - 2g(Sy)

The 2nd equation I'm thinking i should use since it has just one unknown (Uy).

Vy = 0, Sy = 4m

0 = Uy^2 - 2(-9.81)(4)

I don't want to go further until i know what I'm doing so far is correct..

 

[Doc Al]

Ah ok, i thought about the ball not traveling in a straight line assuming the angle was wrong at some point.

Ok so if 'm trying to analyze the vertical component of the initial velocity I am going to use equation:

Sy = Uyt + 0.5(ay)t^2 Or

Vy^2 = Uy^2 - 2g(Sy)

The 2nd equation I'm thinking i should use since it has just one unknown (Uy).

Vy = 0, Sy = 4m

0 = Uy^2 - 2(-9.81)(4)

I don't want to go further until i know what I'm doing so far is correct..

Almost perfect. You have an extra minus sign in your last equation. g is just the magnitude of acceleration, so g = 9.81, not -9.81. (The acceleration is -g = -(9.81) = -9.81 m/s^2.)

 

[Milkster18]

Ok so

Uy^2 = 2(9.81)(4) + 0

Uy^2 = square root 78.48

Uy =8.86 m/s Correct?

What shall i do now?

Do i find the time?

Sy = Uyt + 0.5(9.81)t^

Sy = 4 + (8.86)t - 4.905t^2

Sy = 12.86t / 4.905^2

4 = 12.86 - 4.905

4 = 7.955

t = square root 0.5 = 0.7s

t = 0.7 s .. is this correct so far?

 

[Doc Al]

Ok so

Uy^2 = 2(9.81)(4) + 0

Uy^2 = square root 78.48

Uy =8.86 m/s Correct?

Yes. Good.

Do i find the time?

Sy = Uyt + 0.5(9.81)t^

Sy = 4 + (8.86)t - 4.905t^2

Sy = 12.86t / 4.905^2

Right approach, but you messed up the equation:
(1) 4 is the final position, not the initial.
(2) a + bt ≠ (a + b)t

Even easier: What's the average vertical speed? Use that to find the time.

 

[Milkster18]

Average vertical speed..

Uy / g

8.86 / 9.81 = 0.9 s?

(Sorry I am getting tired as its nearly 2am)

 

[Milkster18]

speed is calculating in m/s.. sorry I am confused

 

[Doc Al]

Average vertical speed..

Uy / g

8.86 / 9.81 = 0.9 s?

No. The vertical speed varies uniformly from its initial value of 8.86 m/s to 0 at the highest point. What do you think the average vertical speed would be?

 

[Milkster18]

sice it highest point is Sy = 4m

is it 8.86 / 4 = 2.215 m/s average vertical speed?

if its not can you provide further help? thanks

 

[Doc Al]

sice it highest point is Sy = 4m

is it 8.86 / 4 = 2.215 m/s average vertical speed?

No. You cannot divide a speed by a distance and expect to end up with a speed.

if its not can you provide further help?

Try this analogy: You start out making zero money and, uniformly, little by little, end up making $100/day. Over that time period, what was your average salary per day?

You may want to review some of the basic formulas of kinematics:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2"
http://www.physicsclassroom.com/Class/1DKin/U1L6a.cfm"

And here's some good stuff about projectile motion:
http://www.physicsclassroom.com/Class/vectors/u3l2a.cfm"

 

[Milkster18]

u+v / 2 = the average?

So v = 0

8.86 / 2 = 4.43 m/s

 

[Doc Al]

u+v / 2 = the average?

So v = 0

8.86 / 2 = 4.43 m/s

Good! So now that you have the average vertical speed and the vertical distance, figure out the time.

 

[Milkster18]

with the analogy:

Say you made 0$ to 100$ over 7 days .. average income per day is 100 / 7 = 14.3$ per day

 

[Milkster18]

Sy = (4.43)t - (4.91)t^2

4 = (4.43)t - (4.91)

4 = -0.48?

4 / 0.48 = square root 8.33

t = 2.887 s ?

 

[Milkster18]

Question: Why does the AVERAGE Initial vertical velocity have to be used instead of just the initial vertical velocity so i can understand why?

 

[Doc Al]

with the analogy:

Say you made 0$ to 100$ over 7 days .. average income per day is 100 / 7 = 14.3$ per day

No. Your income increases from 0/day to 100/day. Like this:
0/day; 1/day, 2/day, 3/day... 100/day. Average (assuming uniform increase--just like uniform acceleration) is (0 + 100)/2 = 50/day. Half the time you made less, half the time you made more.

(But don't worry about that analogy if it doesn't help.)

 

[Doc Al]

Sy = (4.43)t - (4.91)t^2

4 = (4.43)t - (4.91)

4 = -0.48?

4 / 0.48 = square root 8.33

t = 2.887 s ?

No, using that method you would use the initial velocity, not the average velocity. You'll end up with a quadratic to solve.

Question: Why does the AVERAGE Initial vertical velocity have to be used instead of just the initial vertical velocity so i can understand why?

Again, using that method you would use the initial velocity.

I wanted to show you an easier way to get the time, using Distance = ave speed X time. No quadratics to solve. (Do it both ways and compare the answers.)

 

[Milkster18]

Yes i realize using quadratics is a way to find time, i didn't know there was an easier way to solve for t thou.

as for getting the average vertical initial velocity .. i don't know what this answer goes into which equation in order to solve t.

Sorry i haven't solved for initial velocity before as this topic (projectile motion) is new to me.

Easier methods are always better :)

solving for Ux i would have to find time t?

 

[Doc Al]

Yes i realize using quadratics is a way to find time, i didn't know there was an easier way to solve for t thou.

You should solve it both ways to make sure you understand them.

as for getting the average vertical initial velocity .. i don't know what this answer goes into which equation in order to solve t.

I gave you the needed equation in my last post--in words. (Another version is in the first link I gave you.)

Sorry i haven't solved for initial velocity before as this topic (projectile motion) is new to me.

You're doing OK. Keep at it!

Easier methods are always better :)

solving for Ux i would have to find time t?

Yes. To solve for Ux, you'll first need the time. Which is what you're solving for now.

 

[Milkster18]

Ok so in order to find t i use:

Ur saying: Distance = avg speed x time

So time = distance / avg speed?

Got 2 distances sy = 4, sx = 20m

Not sure wat to do here..

Im guessing 4 / 4.43 = 0.9s

Or 4.43/4 = 1.11s

 

[Doc Al]

Ok so in order to find t i use:

Ur saying: Distance = avg speed x time

So time = distance / avg speed?

Yes.

Got 2 distances sy = 4, sx = 20m

Since we just found the average vertical speed, you'll use the vertical distance.

Not sure wat to do here..

Im guessing 4 / 4.43 = 0.9s

Right. (But why guess? Just plug into the formula.)

Or 4.43/4 = 1.11s

That doesn't work. We want distance/speed not speed/distance.

It always helps to check units:

distance/speed = (m)/(m/s) = 1/(1/s) = s (makes sense, since we are trying to find a time)

speed/distance = (m/s)/m = 1/s (no good--1/s is not a proper unit for time)

 

[Milkster18]

Ok, sorry i didn't know about distance = avg speed x time formula. Wasn't told by my lecturer about that.

yes distance / speed sounds better than speed / distance lol.

so now i find Ux given by time.

Sx = Ux t

Ux = Sx / t =

Ux = 20 / 0.9 = Ux as 22.22 m/s

so if I am finding average of Ux:

average horizontal initial velocity = 11.11 m/s?

So now i have the horizontal and vertical component, how do i use these to find the initial velocity?

 

[Doc Al]

Ok, sorry i didn't know about distance = avg speed x time formula. Wasn't told by my lecturer about that.

yes distance / speed sounds better than speed / distance lol.

so now i find Ux given by time.

Sx = Ux t

Ux = Sx / t =

Ux = 20 / 0.9 = Ux as 22.22 m/s

OK.

so if I am finding average of Ux:

average horizontal initial velocity = 11.11 m/s?

(1) Since there's no horizontal acceleration, the average horizontal speed is the same as the initial horizontal speed Ux. (It doesn't vary from 0 m/s to 22.22 m/s; it's constant at 22.22 m/s.)
(2) Ux is what we want, anyway. (We only used average vertical speed as a means to find the time.)

So now i have the horizontal and vertical component, how do i use these to find the initial velocity?

Ux and Uy are the components of the velocity vector. Use Pythagoras to find the magnitude and some trig to find the angle.

 

[Milkster18]

Ok looking good.

Square root 8.86^2 + 22.22^2

=23.9 m/s

Ui = 23.9 m/s .. looking good its correct :)

Any reason to find the angle? I am guessing to help find other components?

If anything ever comes to using quadratic equations is there any other method that is easier such as the average vertical initial component? I've seen quadratic equations involving Sin/Tan/Cos and it looks ever so confusing.

Thanks again Doc Al, before i finish this thread i was just wondering about finding something off topic:

Angular Motion:

Question: A flywheel increases its frequency of rotation from 20rpm to 40 rpm in 10 s with a uniform angular acceleration. Calcualte a) the angular acceleration, and (b) the number of revolutions made during the 10s period.

Answer to a) is 0.21rad/s^2

b) revolutions is 5 ...

but i don't understand how to calculate how many revolutions in a period of time giving the answer to question (b) above

I can calculate everything else apart from the revolutions because once again i wasn't given a formula or teaching of how to find this. Would you be able to provide information on how to find this ?

Thanks again.

 

[Doc Al]

Ok looking good.

Square root 8.86^2 + 22.22^2

=23.9 m/s

Ui = 23.9 m/s .. looking good its correct :)

Good!

Any reason to find the angle? I am guessing to help find other components?

You'd find the angle to fully specify the initial velocity. (If you need to; the question is ambiguous.) You already have the components.

If anything ever comes to using quadratic equations is there any other method that is easier such as the average vertical initial component? I've seen quadratic equations involving Sin/Tan/Cos and it looks ever so confusing.

Sometimes there's no way around it and you have to solve a quadratic.

Thanks again Doc Al, before i finish this thread i was just wondering about finding something off topic:

Angular Motion:

Question: A flywheel increases its frequency of rotation from 20rpm to 40 rpm in 10 s with a uniform angular acceleration. Calcualte a) the angular acceleration, and (b) the number of revolutions made during the 10s period.

Answer to a) is 0.21rad/s^2

b) revolutions is 5 ...

but i don't understand how to calculate how many revolutions in a period of time giving the answer to question (b) above

The basic equations of motion for rotational motion are exactly analogous to the equations for linear motion. Here the number of revolutions (or the angle) is analogous to distance.

One easy way is to use our average velocity trick to find the distance/revolutions. Careful with units.

I can calculate everything else apart from the revolutions because once again i wasn't given a formula or teaching of how to find this. Would you be able to provide information on how to find this ?

Check out this list: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin"

 

[Milkster18]

So use equation

Pheta = W1 + 0.5at^2

W1 = 2.1rad/s
W2 = 4.19 rad/s

Pheta = 2.1(10) + 0.5(0.21)(10)^2

Pheta = 21 + 10.5

Pheta = 30.5 radians?

How do i use this to find revolutions if this is correct?

 

[Milkster18]

31.5 radians.. my mistake

 

[Doc Al]

So use equation

Pheta = W1 + 0.5at^2

W1 = 2.1rad/s
W2 = 4.19 rad/s

Pheta = 2.1(10) + 0.5(0.21)(10)^2

Pheta = 21 + 10.5

Pheta = 30.5 radians?

How do i use this to find revolutions if this is correct?

Good. (Except for the typo, which you corrected later.)

How many radians equal one revolution?

 

[Doc Al]

And don't forget to try the 'easy way'. What's the average angular speed in rpm for that interval?

 

[Milkster18]

one revolution = 2*pi radians

31.5 / 2*pi = just over 5 revolutions :)

but i had to find 2*pi first then use that answer to divided into 31.5 rads.. otherwise the answer will be completely different for some reason.

As for average angular speed:

20 / 2 = 10 rpm?

 

[Doc Al]

one revolution = 2*pi radians

31.5 / 2*pi = just over 5 revolutions :)

Good.

but i had to find 2*pi first then use that answer to divided into 31.5 rads.. otherwise the answer will be completely different for some reason.

Not sure what you mean.

As for average angular speed:

20 / 2 = 10 rpm?

No. It goes from 20 rpm to 40 rpm, so what would the average be? (It can't be 10 rpm!)

 

[Milkster18]

Ah

W1 + W2 / 2

20 + 40 / 2 = 30 rpm

So now i have average how does this make things easier?

Im from the UK by the way so i think some units / symbols are different to US if your from there.

Thanks for the help again. :)

 

[Doc Al]

Ah

W1 + W2 / 2

20 + 40 / 2 = 30 rpm

Good.

So now i have average how does this make things easier?

Use the same formula as before: Distance = ave speed X time
Here "distance" is replaced by angle/revolutions.

Hint: How many minutes is 10 seconds?