Projectile Motion - Initial velocity/total time

[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!

 

[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?

 

[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 v_x= v₀cos(0) and v_y = v₀sin(45) - (9.8m/s²)t ? Is my initial vertical velocity gravity?

 

[rock.freak667]

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 v_x= v₀cos(0) and v_y = 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 v_0x=v₀sin45 but the horizontal component should be v_0x=v₀cos45.

With this in mind and using y=v_0yt-0.5gt², when the object hits the ground, y=0, so what is the time 't' in terms of 'v₀' ?

 

[Anyua]

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

You are correct in saying that the vertical component is v_0x=v₀sin45 but the horizontal component should be v_0y=v₀cos45.

With this in mind and using y=v_0yt-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 v_0y, 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...

 

[rock.freak667]

Ok, so then I am still stuck on the initial vertical velocity, because I would have to use that to find v_0y, 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...

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

v_0y=v₀sin45

and v_0x=v₀cos(45).

If we are considering vertical motion, then y=v_0yt-0.5gt². When the object hits the ground 'y=0', pretend that you know v_0y and solve for 't'. What do you get?

 

[Anyua]

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

v_0y=v₀sin45

and v_0x=v₀cos(45).

If we are considering vertical motion, then y=v_0yt-0.5gt². When the object hits the ground 'y=0', pretend that you know v_0y 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?

 

[rock.freak667]

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 ?

 

[Anyua]

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 = v_0x x t.
81.1 = v₀cos(45) x t
81.1/v₀cos(45) = t?

 

[rock.freak667]

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?

 

[Anyua]

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?

 

[rock.freak667]

That should be correct.

 

[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?

 

[Anyua]

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!