Projectile Motion (Initial Velocity)


by Milkster18
Tags: initial, physics, projectile, velocity
Milkster18
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#1
May4-11, 11:52 AM
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Hi, im 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 im 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
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#2
May4-11, 05:29 PM
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Quote Quote by Milkster18 View Post
So im 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.
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#3
May4-11, 06:30 PM
P: 22
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 im 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 dont want to go further until i know what i'm doing so far is correct..

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#4
May4-11, 06:46 PM
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Projectile Motion (Initial Velocity)


Quote Quote by Milkster18 View Post
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 im 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 dont 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.)
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#5
May4-11, 07:21 PM
P: 22
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?
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#6
May4-11, 07:37 PM
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Quote Quote by Milkster18 View Post
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.
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#7
May4-11, 07:53 PM
P: 22
Average vertical speed..

Uy / g

8.86 / 9.81 = 0.9 s?

(Sorry im getting tired as its nearly 2am)
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#8
May4-11, 07:55 PM
P: 22
speed is calculating in m/s.. sorry im confused
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May5-11, 01:36 PM
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Quote Quote by Milkster18 View Post
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?
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#10
May5-11, 02:01 PM
P: 22
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
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#11
May5-11, 02:18 PM
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Quote Quote by Milkster18 View Post
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:
Basic Equations of 1-D Kinematics
Kinematic Equations and Problem-Solving

And here's some good stuff about projectile motion:
Projectile Motion
Milkster18
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#12
May5-11, 02:36 PM
P: 22
u+v / 2 = the average?

So v = 0

8.86 / 2 = 4.43 m/s
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May5-11, 02:37 PM
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Quote Quote by Milkster18 View Post
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.
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#14
May5-11, 02:43 PM
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with the analogy:

Say you made 0$ to 100$ over 7 days .. average income per day is 100 / 7 = 14.3$ per day
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#15
May5-11, 02:57 PM
P: 22
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 ?
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#16
May5-11, 02:58 PM
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Question: Why does the AVERAGE Initial vertical velocity have to be used instead of just the initial vertical velocity so i can understand why?
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May5-11, 03:21 PM
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Quote Quote by Milkster18 View Post
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.)
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#18
May5-11, 03:26 PM
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Quote Quote by Milkster18 View Post
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.

Quote Quote by Milkster18 View Post
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.)


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