Physics projectile motion question

[Krookwood]

Homework Statement

a gold ball is hit with an initial velocity of 52m/s at an angle of 50 degrees
How long is it in the air

Homework Equations

Sin=opp/hyp
Cos=adj/hyp
D=vt+1/2at^2

The Attempt at a Solution

First I split the vector into its vertical and horizontal components
sin50=x/52
52sin50=x
x=39.83m/s (up)

cos50=x/52
52cos50=x
x=33.4m/s (Horizontal)

Then I use the kinematic equation to find the amount of time it is in the air considering its total vertical displacement will be 0
d_v=v₁*t+1/2a*t²
0=(39.83m/s)(t) + 1/2 (-9.8m/s²)t²
i then factor out "t" and this is where I'm having an issue. I get my equation to look like this
0=t(39.83m/s - 4.9m/s² t)

and the supposed answer is "0 or 8.1" however I'm unable to get this same result, I may be forgetting something so if anyone could explain how I get this number I'd greatly appreciate it.

 

[Dr. Courtney]

You need to take care to express both the vertical and the horizontal equations of motion and to distinguish them from each other. Your symbols are also confusing position and velocity.

 

[Gianmarco]

Homework Statement

how far will a football travel when it is thrown at 62km/h at an angle of 35 degrees to the ground.

Homework Equations

Sin=opp/hyp
Cos=adj/hyp
D=vt+1/2at^2

The Attempt at a Solution

First I split the vector into its vertical and horizontal components
sin50=x/52
52sin50=x
x=39.83m/s (up)

cos50=x/52
52cos50=x
x=33.4m/s (Horizontal)

If your $V_0 = 62\frac{km}{h}$ and your angle is 35°, why are you using $V_0 = 52\frac{km}{h}$ and an angle of 50°

 

[Krookwood]

sorry i wrote the wrong question I've corrected it

 

[Gianmarco]

sorry i wrote the wrong question I've corrected it

I also think that the question isn't about how far the ball will travel, but after how much time it will hit the ground

 

[Krookwood]

yeah sorry, I have a lot of work to do and I'm making mistakes. sorry.

 

[Gianmarco]

It's okay, just look at your result

d_v=v₁*t+1/2a*t²
0=(39.83m/s)(t) + 1/2 (-9.8m/s²)t²
0=t(39.83m/s - 4.9m/s² t)

and think of what the problem is asking you

 

[Krookwood]

I'm having severe math block and honestly, can't figure it out. This is a support question for an online class so after you give them the answer they give you the answer and the correct steps to follow, it takes about 3 days for a teacher to reply so essentially I have no one to ask. everything is right for me until this point, and I'm unsure what jump they made to get "0 or 8.1"What steps do you use to add 39.83m/s with -4.9m/s²

 

[Gianmarco]

You want to find the time the ball will take to hit the ground during its motion. The equation for displacement along the y-axis is $y(t) = v_yt-\frac{g}{2}t^2$. By setting $y(t) = 0$, like you did, you will find the time the ball takes to hit the ground. $0 = t(v_y - \frac{g}{2}t)$. Both sides of the equation should be zero, so you have to find the values of t(it's a quadratic equation, so you expect two solutions) for which the right side is zero.

 

[Krookwood]

Thanks so much for the help, i don't know how i didn't realize that i needed the quadratic equation especially considering i had two possible results given.

 

[inderr]

Its easy. Just do 52mph * 50/ph (angle), and you get 50*52 seconds which is 8 minutes

source: pHd in kemistry at harverd