# Projectile Fired Upward

1. Jan 30, 2004

### KingNothing

I am not sure how to do this. I keep coming out with the wrong answer.

"a projectile is fired straight up at an initial velocity of 58.8 m/s. How high above the ground is it 1.5 seconds after firing?"

2. Jan 30, 2004

### Rog

You have not supplied enough information. You would need to know the size, mass of the object to factor in wind resistance and kentic energy along with the effect of gravity.

3. Jan 30, 2004

### Staff: Mentor

This should be in the homework help section.

Since you didn't show your work, we can't tell where you went wrong.

Ignoring the complication of air resistance, this is a straightforward application of the kinematics of constant acceleration:
$$y = y_0 + v_0t + \frac{1}{2}at^2$$

4. Jan 30, 2004

### Staff: Mentor

For the sake of a high school physics problem (which this appears to be), none of that is relevant (except the effect of gravity, which is known).

I always split these problems into two parts for easier swallowing. First, use the deceleration due to gravity to find the speed of the projectile after 1.5 sec. Then multiply the time by the average speed.

5. Jan 30, 2004

### Staff: Mentor

Nothing wrong with that. For many students, it's the best approach.

I encourage students who are comfortable with the algebra to know (or derive!) the relationships between the three variables: distance, time, and speed. One should be able to produce at will a formula connecting any two of these variables (for constant acceleration). And understand how to apply them.

6. Jan 30, 2004

### KingNothing

Yeah, I think thats the approach I ended up taking (I had to do this problem today, but sadly I didn't get to check this forum in time to see if there were any replies.)

Now this is really jsut pissing me off. I can't remember how I did it. AHHH!!!!! I hate that feeling. I want to say what I got for an answer to see if you all think its reasonable, but I can't even remember. If its not too much trouble, could someone give me a good or reasonable answer? I don't know why my memory chose to fail me now.

Scratch the above idea...let me see if I can remember.

What is y in that equation?

Last edited: Jan 30, 2004
7. Jan 30, 2004

### Jimmy

y is referring to distance or displacement. $y_0$ is the initial displacement, which in this case is zero.

Last edited: Jan 30, 2004
8. Jan 30, 2004

### KingNothing

I got D=77.175m. Can anyone confirm this?

9. Jan 30, 2004