# Homework Help: Projectile Motion and Velocity Problem

1. Oct 3, 2008

### maniacp08

A rock is thrown from the top of a 26 m building at an angle of 51° above the horizontal.

(a) If the horizontal range of the throw is equal to the height of the building, with what speed was the rock thrown?

(b) How long is it in the air?

(c) What is the velocity of the rock just before it strikes the ground?

Let Vi = initial velocity
X component = Vi Cos 51
Y component = Vi Sin 51

I know that without any height, I can calculate the time
0 - Vi Sin 51 = -9.81 * T
T = Vi Sin 51 / 9.81

Total time in the air = 2 Vi Sin 51 / 9.81

Horizontal range = Vix * Total time in air
26 = Vi Cos 51 * 2 Vi Sin 51 / 9.81

but since there is a height involved, what changes do I need to make?

2. Oct 3, 2008

### Hootenanny

Staff Emeritus
When the rock is thrown off the top of the building what is the rocks final vertical position relative to the top of the building?

3. Oct 3, 2008

### maniacp08

Im not too sure on the question but if a rock is thrown off the top of the building, the final vertical position is another 26 meters down?

4. Oct 3, 2008

### Hootenanny

Staff Emeritus
Correct. So you know the final vertical displacement (-26m), the acceleration, intitial vertical velocity and you want to find the time of flight. Can you think of a kinematic equation with all those variables in?

5. Oct 3, 2008

### maniacp08

It would be
D = Vi * T + 1/2 A * T^2

But I don't know Vi vertical.
Vi Vertical is Vi Sin 51, The Vi is still unknown.

6. Oct 3, 2008

### Hootenanny

Staff Emeritus
Correct, but you do know the vertical range. Can you form another equation using this information?

Don't forget that it should be Vi sin(51) not just Vi

7. Oct 3, 2008

### maniacp08

Isn't the vertical range formula the same for horizontal range except the component differences?

Vertical range formula is
Dy = Yo + Viy * T

but Viy = Vi sin(51) we don't know Vi tho.

I can't think of any more formula that utilizes these info.

8. Oct 3, 2008

### Gib Z

The formula's are in fact not the same - The horizontal component has no acceleration, whilst the vertical component is influenced by Acceleration due to Gravity. You know when it has reached the peak of its vertical motion, the y-component of its velocity is equal to ... ? You also know the displacement in the y- direction and acceleration in the y-direction. You want to find the Initial velocity in the y-direction . Try to remember a formula that relates these variables.

9. Oct 4, 2008

### maniacp08

Ahh, is it
Vf^2 = Vi^2 + 2 * A * D?

Vi^2 = 510.12
Vi = 22.5 m/s

Hmm, I gave this a little thought, Im not sure if I am correct.
I have the initial vertical velocity now so I will use

Vf = Vi + A * T
where Vi = 22.5 sin 51 and A = 9.81
Where Vf = 0 to find the time that it hits the highest point
Where T = 1.8s

Using the time that hits the highest point I plug it in
D = Vi * T + 1/2 A * T^2
to find the distance it traveled
and D = 47m up

Since it reached 47m up, it must come down another 47m to its initial position thrown
which is 94m but since it falls to the ground that is another 26m down from the buildings height
which D = 120m total.

Now I use
D = Vi * T + 1/2 A * T^2
where D = 120 to find the total time in the air
T = 3.5s

Plug that in the horizontal range equation
D = Vix * T
120 = Vi Cos 51 * 3.5
Vi = 12 m/s

Am I Right?
I hope so!!

Can someone help me with this
(c) What is the velocity of the rock just before it strikes the ground?

Last edited: Oct 4, 2008
10. Oct 4, 2008

### Hootenanny

Staff Emeritus
That would be indeed the correct formula, but where have you got that initial velocity from?

11. Oct 4, 2008

### maniacp08

Im srry I made an edit on my post, could you please look at it.

12. Oct 4, 2008

### Hootenanny

Staff Emeritus
You still haven't said how you obtained the initial vertical velocity.

13. Oct 4, 2008

### maniacp08

I use the formula that you deemed correct and plug in the numbers.

Vf^2 = Vi^2 + 2 * A * D
Vf = 0 because it is at its highest position
D = 26m

to find Vi. which is 22.5 and Initial vertical velocity is 22.5 sin 51.

14. Oct 4, 2008

### Hootenanny

Staff Emeritus
The problem is that the velocity is non-zero when the displacement is 26m. When the ball is at y=26, the velocity is equal to the initial velocity.

15. Oct 4, 2008

### maniacp08

Ok now Im confused.
So the final velocity is not 0 when is at y=26 and is equal to its initial? So how would I approach this?

16. Oct 4, 2008

### Hootenanny

Staff Emeritus
Yes, because at y=26m the ball is on top of the building. As I said before, use the equation in your above post and fill in all the variables that you know. You will be left with two unknowns, hence you will need to construct a further equation using the information you are given in the question.

17. Oct 4, 2008

### maniacp08

Oh so what I did before was wrong then.

Then we are back to square 1 now but even worse, we have 2 unknowns.
Vf^2 = Vi^2 + 2 * A * D
Vf^2 = Vi^2 + 2 * -9.81 * 26

Can you throw me a hint?

18. Oct 4, 2008

### Hootenanny

Staff Emeritus
Consider the vertical motion first. Try using this equation that you posted before:
Fill in all the information you know.

19. Oct 4, 2008

### maniacp08

D = Vi * T + 1/2 A * T^2

26 = Vi * T + 1/2(9.81) * T^2

Unknowns are Vi and T

20. Oct 4, 2008

### Hootenanny

Staff Emeritus
Close, but don't forget that the final position of the ball is below it's initial position and therefore D=-26m. Further, the acceleration due to gravity is in the negative y direction and therefore a=-9.81 m/s2.

So you have two unknowns, Vi and T. Can you now write down an equation that governs the horizontal motion?

21. Oct 4, 2008

### maniacp08

The horizontal range equation?
Dx = Vix * T

26m = Vi cos 51 * T
unknowns are Vi and T

22. Oct 4, 2008

### Hootenanny

Staff Emeritus
Sounds good to me. So you now have two equations:

$$-26 = v_i\sin(51)t - \frac{1}{2}gt^2$$

$$v_i\cos(51) = \frac{26}{t}$$

with two unknowns. All that's left to do is solve them.

23. Oct 4, 2008

### maniacp08

oooo!!!!!!
I solved t first and found that vi = 26/T cos51
and I got T = 3.44s

and got Vi = 12m/s
26 = Vix cos 51 * 3.44

Thanks for your help! it was a tough one.
Is there any advice on how to approach these projectile problems?
The only reason I solved this was because of your help, my exam is coming up and I don't know how to approach these problems.

24. Oct 4, 2008

### Hootenanny

Staff Emeritus
My pleasure.

Well, the way I would usually recommend to approach such problems is to first split the motion into horizontal and vertical components, then make a list of all the knowns. If you have two unknowns, then you know that you need to find two appropriate kinematic equations; if you have three unknowns, then you need to find three equations.

It is also always a good idea to define your coordinate system at the outset, even if it's just with a sketch. This makes it easier to see which quantities should be positive and negative and helps avoid silly sign slips.

25. Oct 4, 2008

### maniacp08

Ok, I decided to put what I learned from this problem on to another problem.

An archer fish launches a droplet of water from the surface of a small lake at an angle of 60 degrees above the horizontal. He is aiming at a juicy spider sitting on a leaf 50 cm to the east and on a branch 25cm above the water surface. The fish is trying to knock the spider into the water so that the fish can eat the spider.

A) What must the speed of the water droplet be for the fish to be successful?
B) When it hits the spider is the droplet rising or falling?

We are given the horizontal range which is 50cm
and the vertical range which is 25

So I separated them into 2 components
the vertical I have
25 = Viy sin 60 * T + 1/2 G * T^2
where Viy and T are unknowns

The horizontal component
Dx = Vix Cos 60 * T
50 = Vix Cos 60 * T

Unknowns are Vix and T
I try doing the same thing solving for either Vi or T and plug it in.
Is this approach correct for this problem?