# Homework Help: Projectile Motion - Initial velocity/total time

1. Sep 16, 2010

### Anyua

The problem:
The record distance in the sport of throwing compats is 81.1m. This record toss was set by Steve Urner of the US in 1981. Assuming the initial launch angle was 45 degrees and neglecting air resistance, determine a) the initial speed of the projectile and b) the total time the projectile was in flight.

What I have so far:
I am still a bit uncertain as to how initial velocities work in 2 dimensions, but I think, since it is along the x-axis, then the initial velocity is equal to 0...? And the second part seems so simple but I can't find an equation to use that I have enough numbers for. I am not sure if I need to solve for the maximum height first then solve for time, or if I can do it by manipulating another equation? Not really sure either where the angle comes in, unless I am doing a maximum height first...?

Thanks for any help or tips!

2. Sep 16, 2010

### rock.freak667

Let's say the initial speed is v0 which is thrown at 45°. What are the x and y components of this velocity?

Now the x-component/horizontal velocity stays constant.

They said that the object traveled 81.1 m, meaning that the horizontal distance traveled is 81.1.

Horizontally, distance = speed*time. So formulate this expression.

A time function is present which we do not have. We know that x=v0t-0.5gt2.

So vertically, when the object hits the ground the vertical displacement is zero.

So vertically, what is your initial velocity?

3. Sep 16, 2010

### Anyua

Hmm...I thought it was y=v0t-0.5gt2...? I am really bad at manipulating these formulas to work for me :( the x and y components are vx= v0cos(0) and vy = v0sin(45) - (9.8m/s2)t ? Is my initial vertical velocity gravity?

4. Sep 16, 2010

### rock.freak667

sorry, I used the same symbol twice, it should be y=v0t-0.5gt2.

You are correct in saying that the vertical component is v0x=v0sin45 but the horizontal component should be v0x=v0cos45.

With this in mind and using y=v0yt-0.5gt2, when the object hits the ground, y=0, so what is the time 't' in terms of 'v0' ?

Last edited: Sep 16, 2010
5. Sep 16, 2010

### Anyua

Ok, so then I am still stuck on the initial vertical velocity, because I would have to use that to find v0y, right? I am still thinking it's 0. If so, I would turn the problem around to t=$$\sqrt{}-4.9 m/s2\sqrt{}$$ but that is wrong...argh...

6. Sep 16, 2010

### rock.freak667

Let's start over, I may have confused you, we know the components are

v0y=v0sin45

and v0x=v0cos(45).

If we are considering vertical motion, then y=v0yt-0.5gt2. When the object hits the ground 'y=0', pretend that you know v0y and solve for 't'. What do you get?

7. Sep 16, 2010

### Anyua

Haha, ok - so I would have y, which = 0, = v0sin(45)t - (4.9 m/s2)t2. Then v0sin(45)t = (4.9 m/s2)t2 since my other side is equal to 0. Then v0sin(45)t/(4.9 m/s2)t2 and multiply out my single t to get v0sin(45)/4.9 m/s2 --- is that right?

8. Sep 16, 2010

### rock.freak667

Yes that is good, so the time for the entire motion is t=v0sin(45)/4.9 m/s2.

Now let's consider horizontally.

The horizontal component is not affected by gravity, so it stays constant, giving the horizontal distance traveled as horizontal velocity*time.

So what is the horizontal distance equal to ?

9. Sep 16, 2010

### Anyua

81.1 = v0x x t.
81.1 = v0cos(45) x t
81.1/v0cos(45) = t?

10. Sep 16, 2010

### rock.freak667

Good! And from before we have t=v0sin(45)/4.9, so if we sub for t, can we get v0?

11. Sep 16, 2010

### Anyua

Ok, so:

v0sin(45)/4.9m/s2 = 81.1m/v0cos(45)
4.9m/s2/v0sin(45) * 81.1m/v0cos(45)
v02 = 4.9m/s2/sin(45) * 81.1m/cos(45)
v0 = 28.2m/s?

12. Sep 16, 2010

### rock.freak667

That should be correct.

13. Sep 16, 2010

### Anyua

And then I'd plug that back into y = 28.2 m/s * sin(45) - 9.8 m/s^2t?
t= (28.2)(sin45)/9.8 = 2.03s?

14. Sep 16, 2010

### Anyua

Ok! Oh my goodness, thank you sooo much for your help! Sorry for getting so confused, this is my first semester taking physics ever! Luckily I got the college physics course without the calc, haha. Thank you!!!