# Homework Help: How Can I Find Displacement With Only Velocity and Angle

1. Sep 13, 2015

### noname2020x

1. (Ignore air resistance)
A rock is thrown 55% above the horizon at 30m/s. What is the horizontal displacement of the rock when it hits the ground?

2. Can only use vf^2=vi^2+2ad and sohcahtoa

3. Would it work to first draw a vector triangle and find the vertical velocity vector only.
Then use vf^2=vi^2+2ad to find the displacement of the rock (vertically) by using vf=0 and vi as the vertical velocity calculated and -9.8 as acceleration.
Then calculate horizontal displacement and multiply by 2?

Thanks!

2. Sep 13, 2015

### BOAS

This question, as you have correctly deduced is essentially one of "hang time".

For how long is the rock in the air, allowing it to travel horizontally?

3. Sep 13, 2015

### noname2020x

Cool.So I'm going about solving it the right way?

Also, in the beginning I didn't mean % I meant degrees haha.

4. Sep 13, 2015

### BOAS

From what you have said, I think you're on the right tracks. Post your working if you wish me to take a look.

5. Sep 13, 2015

### noname2020x

Here is the basic outline.

#### Attached Files:

• ###### 20150913_135924.jpg
File size:
50.9 KB
Views:
142
6. Sep 13, 2015

### SteamKing

Staff Emeritus
In projectile motion, the initial launch angle does not remain constant, because the projectile's vertical velocity is being reduced by the effect of gravity.

What you want to do is find out how long it takes after launch for the projectile to stop rising vertically, and begin to fall back to earth.

7. Sep 13, 2015

### noname2020x

That is what the vf^2=vi^2+2ad is for. I am isolating the vertical velocity vector to find vertical displacement.

8. Sep 13, 2015

### SteamKing

Staff Emeritus
You don't really need to calculate the vertical displacement, since you are really interested in calculating the horizontal displacement.

Finding the time the projectile remains aloft will allow you to calculate what the horizontal displacement is, however.

9. Sep 13, 2015

### noname2020x

How might that be accomplished?

Wouldn't this method still work?

10. Sep 13, 2015

### SteamKing

Staff Emeritus
Just think about what knowing the total time aloft means for figuring the horizontal travel distance of a projectile. If you know the total time aloft and the horizontal velocity of the projectile, then finding the distance traveled is a simple multiplication (assuming no resistance from the atmosphere).
It probably would, but you must do a lot of pointless calculations to reach the final result. It's like traveling sideways when you really should push straight ahead to reach your goal.

There are many example projectile calculations which you can find to study on the web (and on this site).

11. Sep 23, 2015

### noname2020x

If I do it with time the distance in the x-direction is twice as long (and actually accurate according to simulations)

Why isn't it working with my original calculations?

12. Sep 23, 2015

### SteamKing

Staff Emeritus
I don't know. I asked you to post your calculations, but I don't think I ever got to see your work.

One image you did post showed an initial velocity of 20, but a complete set of calculations was lacking.

This problem calls for the rock to be thrown with a velocity of 30 m/s at an angle of 55° above the horizon.