# Homework Help: Question for projectile motion

1. Oct 2, 2015

### GZM

1. The problem statement, all variables and given/known data
The question states :
Romeo tosses a pebble at Juliet's window to wake her. Unfortunately, he throws too large a pebble too fast. Just before crashing through the glass, the pebble is moving horizontally, having traveled a horizontal distance x and a vertical distance y as a projectile

Find the magnitude of the pebble's velocity as it leaves Romeo's hand.

2. Relevant equations
Unsure, thinking of something like kinematic equations

3. The attempt at a solution
It says just before crashing the glass the pebble is horizontal, I related that fact to a parabolic trajectory where the apex of the upside down parabola is where the pebble breaks the glass, I have 2 known things gravity = 9.8m/s^2 and the final velocity at apex is 0m/s in the Y axis, I am unsure what other information I can use and what equations to use to solve this problem please help!

2. Oct 2, 2015

### 24forChromium

you need to know the height of the window. After that, find the initial vertical velocity using some of the equations in your data table. The initial horizontal velocity is the same.

3. Oct 2, 2015

### GZM

hi, but since the height of the window was never given how do i go on about figuring it out?

4. Oct 2, 2015

### 24forChromium

Sorry I misread distance with velocity; the horizontal velocity is constant, so just x/t will do the trick for h_speed. For the vertical velocity it's a bit more complicated, have you learned calculus? If not, use this equation: d = v1 * t + 0.5 * a * t^2, where d is displacement (vector, with direction) and v1 is initial velocity (with dir. again) and a is acceleration (in your case is in the opposite dir. as the velocity) Then some algebra will do the rest. Hope that helped.

5. Oct 2, 2015

### GZM

sorry, I'm not understanding this concept you say use this equation "d = v1 * t + 0.5 * a * t^2" but within that equation I have 3 unknowns, the displacement(which can be said to be Y, but that is an unknown distance introduced in the question), initial velocity and also time, even if I attempt with some algebra I cannot solve for my initial velocity. Is there some other equation I can equate to so i can eliminate or solve some variables?

6. Oct 3, 2015

### 24forChromium

Will you please tell me the variables that is given?

7. Oct 3, 2015

### Staff: Mentor

The problem is not looking for a numerical answer. It is looking for a symbolic answer (a formula).

This is a projectile problem with horizontal and vertical components. Start with the vertical motion. What do you know about it?

Note that you should write out your kinematic equations as Relevant Equations. This is good for practice, memorization, and inspiration when you're looking for an idea for a way to proceed on a problem.

8. Oct 3, 2015

### GZM

Hi there, so when I'm thinking about the problem I can say Vxf = Vxi because horizontal is constant, then for the Y velocity it is : Vyf = Vyi + at, which becomes Vyi = gT/2 because final velocity is 0 at the top and time is also half.
Since the question was asking for magnitude of the pebble leaving Romeo's hand it should addition of vectors which Vxi + Vyi = Vi, substitute Vxi + (gt/2) = Vi would that be considered an answer?, even though there is a part 2 that asks for angle and how am I supposed to figure that out without real numbers?

9. Oct 4, 2015

### Staff: Mentor

Um, why is the time half? What is T? So far you've just introduced new variables.
That would not be a suitable answer as it doesn't relate in any way to the given values x and y, and that is not how vectors add. While you are not given numerical values for x and y here, it is clear from the problem statement that x and y are the two "givens" that you are to work with. You are to treat them as given values.

Your results should be expressions involving the given and known quantities only: x,y, and g.

I suggest that you begin with finding the initial vertical velocity using either energy conservation principles or a handy kinematic equation involving speed, acceleration, and distance. Use another equation to write an expression for the time from launch to impact. Then tackle the x-direction.

Once you have expressions for the components of the initial velocity you should be able to write an expression for the launch angle too.

10. Oct 4, 2015

### GZM

So this is how i re-approached this,
1) i started with the vertical component and used the equation d(y)=Vi*t+(1/2)a*t^2, solved for Vi it became Vi= (d-(1/2)a*t^2)/t
2) i used another equation d= ((vi+vf)/2)t, solved for t = (2d)/vi and vf = 0 at apex of the parabola
3) put equation 2 into equation 1, the final equation as a result is : Vi=Y(Vi-a)/2
4)Horizontal is constant so it would be a simple Vx=d/t so i solved for t which gives me X/Vx
5) add the vectors sqrt(Viy^2+Vx^2) and the answer is : sqrt((Y(Viy-a)/2)^2 + (X/t)^2)

Have i done this correctly? or am i missing something still?

11. Oct 5, 2015

### Staff: Mentor

Your result still contains additional unknowns Viy and t, so it is not a suitable result.

Let's take a slightly different approach by working with a similar problem that has numbers as givens. Suppose you are given the following problem: A stone is thrown vertically from the ground and reaches a height of 6 m, the top of its trajectory. At what speed was the stone launched?

12. Oct 5, 2015

### Mister T

Use a kinematic equation to find the time it would take the stone to fall from a height y. (You can do as gneill suggests and temporarily let y equal 6 m.)

13. Oct 5, 2015

### GZM

Oh I see what you're trying to say I use the kinematic formula of Vf^2=Vi^2+2ad and i solved for Vi, which gave me Vi = sqrt(19.6Y) does that look right? how should I proceed with the horizontal component without keeping time in?

14. Oct 5, 2015

### Staff: Mentor

That's the idea. You should leave the acceleration as the symbol "g" rather than substituting in a number. Do all the work as symbols and constants.

Once you have an expression for the initial vertical velocity, can you find the time the stone takes to reach its apogee?

Last edited: Oct 5, 2015
15. Oct 5, 2015

### Mister T

What you're calling vi should more properly be referred to as viy, because it's the y-component of vi.
The kinematic equation (the one without time) for the horizontal component would look like this:

vx2 = vix2 + 2axΔx.

But that's of no immediate help! So you must use at least one of the kinematics equations that do involve time.

16. Oct 5, 2015

### GZM

I see, so i use another equation, vf = vi + a*t, then i could plug in the value of Vi i found into the equation and I can solve for t in this new equation that would give me
t= sqrt(2gy)/g would that be correct? then I could put this in a simple Vx=d/t constant acceleration equation which would give me Vx=(d*sqrt(g))/sqrt(2y)? then i could just do a simple vector addition of sqrt(ViY^2+ViX^2) = Vi?

Edit : which gives me the answer : sqrt((d^2g+4gy^2)/2y) would this be an appropriate and correct answer?
edit 2 :it is confirmed right answer, thanks for all the help I appreciate it and now to find the angle i just use tan theta = Viy/Vix?

Last edited: Oct 5, 2015
17. Oct 5, 2015

### Staff: Mentor

Well done

Yup, that's how to get the angle.

18. Oct 5, 2015

### Mister T

Your answer is in terms of d, which is not given in the statement of the problem.

19. Oct 5, 2015

### kphysicsplz

not sure how to get rid of the d...

20. Oct 6, 2015

### Staff: Mentor

Well, what was "d" when you introduced it? Remember, the only values your were "given" were x and y.

21. Oct 6, 2015

### Mister T

He's not the OP!

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