Projectile Motion - Initial velocity/total time

In summary, the record distance for throwing compats is 81.1m set by Steve Urner in 1981. Assuming an initial launch angle of 45 degrees and neglecting air resistance, the initial speed of the projectile is approximately 28.2m/s and the total time the projectile was in flight was 2.03 seconds. The horizontal distance traveled is also 81.1m.
  • #1
Anyua
8
0
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!
 
Physics news on Phys.org
  • #2
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
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
Anyua said:
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?

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:
  • #5
rock.freak667 said:
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 v0y=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' ?

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=[tex]\sqrt{}-4.9 m/s2\sqrt{}[/tex] but that is wrong...argh...
 
  • #6
Anyua said:
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=[tex]\sqrt{}-4.9 m/s2\sqrt{}[/tex] but that is wrong...argh...

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
rock.freak667 said:
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?

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
Anyua said:
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?

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
rock.freak667 said:
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 ?

81.1 = v0x x t.
81.1 = v0cos(45) x t
81.1/v0cos(45) = t?
 
  • #10
Anyua said:
81.1 = v0cos(45) x t

Good! And from before we have t=v0sin(45)/4.9, so if we sub for t, can we get v0?
 
  • #11
rock.freak667 said:
Good! And from before we have t=v0sin(45)/4.9, so if we sub for t, can we get v0?

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
That should be correct.
 
  • #13
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
rock.freak667 said:
That should be correct.

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!
 

1. What is projectile motion?

Projectile motion refers to the motion of an object that is launched into the air and then moves under the influence of gravity alone.

2. What is initial velocity?

Initial velocity is the velocity at which an object begins its motion. In projectile motion, it is the velocity at which the object is launched into the air.

3. How is total time calculated in projectile motion?

Total time in projectile motion is calculated by finding the time it takes for the object to reach its maximum height and then doubling that time. This is because the object takes the same amount of time to reach its highest point as it does to fall back down to its initial height.

4. What factors affect the initial velocity in projectile motion?

The initial velocity in projectile motion is affected by the angle of launch, the initial height, and the presence of air resistance. Changing any of these factors will result in a different initial velocity.

5. How does changing the initial velocity affect the total time in projectile motion?

In projectile motion, changing the initial velocity will affect the total time in two ways. A higher initial velocity will result in a longer total time because the object will travel further before reaching its maximum height. However, a higher initial velocity will also result in a shorter total time because the object will be moving faster and will reach its maximum height more quickly.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
206
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
790
  • Introductory Physics Homework Help
Replies
6
Views
392
  • Introductory Physics Homework Help
Replies
11
Views
668
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
Replies
3
Views
3K
Back
Top