# Homework Help: Projectile no air resistance

1. Nov 26, 2009

### 1irishman

1. The problem statement, all variables and given/known data
An object is thrown horizontally from the top of a cliff at a velocity of 20m/s.
If the object takes 4.20s to reach the ground, what is the range of the object?
I got the first question figured out, but i can't figure out the second question which is:
What is the velocity of the object when it hits the ground? This is the one I can't seem to figure out.

2. Relevant equations
v=d/t horizontal uniform motion equation
d= vit +1/2at^2 vertical uniformly accelerated motion equation

3. The attempt at a solution
I got the range to be 84m in terms of the time and velocity given for the answer to the first question that applies to the horizontal distance.

The final velocity when the object hits the ground will be zero I figure. The initial velocity on the vertical will be zero too right? Help?

2. Nov 26, 2009

### mgb_phys

No - have you never dropped anything fragile and expensive? It certainly doesn't hit the ground with zero velocity.

The solution is to calculate the vertical velocity and horizontal velocity separately and then find the magnitude. What do you know about the change in velcoity in both directions?

3. Nov 26, 2009

### 1irishman

This is what I got:

vertical distance:
d=1/2at^2
1/2X9.8X4.2^2=86.2m
-------------------------
then final vertical velocity is:
0=20^2+2X9.8X86
=400+1686
=sqrt 2086=45.7m/s is final velocity before hits the ground
they have in the book 45.8m/s before hits the ground at 64.1deg below the horizontal
I had 45.7m/s at 66deg below horizontal tan-1 45.7/20

4. Nov 26, 2009

### mgb_phys

That's the first step - you have the final vertical velocity.
Now you have to include the horizontal velocity.
Whats the final horizontal velocity?

5. Nov 27, 2009

### 1irishman

horizontal velocity is zero no?

6. Nov 27, 2009

### 1irishman

We can only use formulas the teacher has introduced us too. The one you show here we've never seen, but thanks.

7. Nov 27, 2009

### modulus

Actually, it's quite logical. The horizontal velocity will not change, because there is no acceleration on the x- axis.

8. Nov 27, 2009

### modulus

So, for the x-axis, initial veloctiy=final velocity. You already know the final velocity on the y-axis. Add the componenets from what you've learned in vectors, and you'll have the answer.

9. Nov 27, 2009

### 1irishman

yup you're right, thanks!

10. Nov 27, 2009

### 1irishman

Since Vx = 20 m/s, Vy = g*t = -9.80 * 4.20 = -41.2 m/s

Therefore, the speed of the object is |V| = sqrt ( Vx^2 + Vy^2 ) = sqrt ( (41.2)^2 + 20^2) = 45.8 m/s

The direction of velocity is : tan (theta) = Vy / Vx = -41.2 / 20 = - 2.06 ------> theta = tan-1( -2.06 ) = - 64.1 negative means below horizontal.

That's just so much easier!

11. Nov 29, 2009

### modulus

Great, you got it completel right! You just cleared up a really important concept of yours. The fact that there is no acceleration on the x-axis is often used to derive the majority of the equations you study. Using this knowledge you can now start deriving formulas of your own. Try doing this one:

projectile is launched from the endpoint of this building:
___________
l l
l l
l l
l l
l l
l l
l l
l l So, it will land on the ground somwhere around here
________________________________________________________________________

Figure out the total orizontal range. (Hint: figure out the total time taken and multiply it by the component of the initial velocity on the x-axis.

P.S.: Thanks for saying 'thanks'! It feels good to know that you've helped the othe person!