Calculate time traveled from launch to land

[://Justice]

Homework Statement

So, this is pretty easy and simple, I am just missing something obvious here I am pretty sure. This stems from a question I asked earlier tonight.
Calculate the time it took from launch to land, given a velocity of 30.197m/s, and a distance traveled of 85m. (other irrelevant data: Object was initially launched at 33 degrees. Refer to https://www.physicsforums.com/showthread.php?p=2890328#post2890328 for more info)

Homework Equations

Perhaps, X=Xo+Vot+.5at^2 ?

The Attempt at a Solution

So, I would use the formula X=Xo+Vot+.5at^2, correct? But then, I get a bit confused...
-.5t^2=Vot-X
divide by t
-.5t=Vo-X
t=-2(Vo-X)
but that is incorrect, I think... SO I dunno... Help, please... Thank you

Oh, wait... Am I maybe just way over-analyzing this? Would it be as simple as Distance/Velocity=Time? Therefore, V=30.197m/s , and D=85m, so 85/30.197=2.815s. But that seems a bit short. The empirical data is 4s (we must find the percent error), that is quite a large percent error...

 

[rl.bhat]

In the horizontal direction there is no acceleration.

s = vo*cos(theta)*t.

 

[://Justice]

In the horizontal direction there is no acceleration.

s = vo*cos(theta)*t.

Which is why I excluded 'a' when simplifying the equation, because a=0. Let me try that equation. Hmm... that seemed to yield a much more reasonable answer, 3.356s. That seems to be correct. Yeah, I get it now. It makes sense. Funny how things can just click and then you get it. Thank's for the help, I knew that it was really simple, I was just making it complicated.

Do you think you could help me really quickly with one more? Sorry.
Q: Find the initial velocity, when the object is shot at 70 degrees and took 6.5s to land. (hint: look at the y direction first this time)

Now, I know that there is the method of dividing it up into components of x and y, however I did not quite understand that. When calculating the initial velocity, given distance traveled and degree, I used Range = v02sin2θ/g which worked great (refer to https://www.physicsforums.com/showthread.php?p=2890328#post2890328 ), however, I don't know if that can apply to this too... Sorry, but physics is really confusing me at the minute

 

[rl.bhat]

Q: Find the initial velocity, when the object is shot at 70 degrees and took 6.5s to land. (hint: look at the y direction first this time)

If it lands on the ground, vertical displacement y is zero. Vertical component of the velocity is vo*sinθ.
Use the equation
y = vo*sinθ* - 1/2*g*t^2

Substitute the values and find vo.

 

[rwisz]

I would approach this problem by first identifying the known variables: velocity and distance traveled. Then, I would use the equation D = Vt to calculate the time traveled. In this case, D = 85m and V = 30.197m/s, so t = 85m / 30.197m/s = 2.815s. However, it is important to note that this calculation assumes a constant velocity, which may not be the case in real-world scenarios.

To account for any changes in velocity, we could use the equation D = Vot + 1/2at^2, where a is the acceleration. This would give us a more accurate measurement of the time traveled.

In terms of the percent error, it is important to consider any external factors that may have affected the actual time traveled, such as air resistance or changes in velocity. These could account for the difference between the calculated time and the empirical data. As scientists, we must always be aware of potential sources of error and try to minimize them in our calculations.