# Calculate time traveled from launch to land

• ://Justice
In summary, the conversation is about calculating the time it took for an object to travel from launch to land, given its velocity and distance traveled. The equation X=Xo+Vot+.5at^2 is suggested, but the person gets confused and wonders if it could be as simple as Distance/Velocity=Time. The equation s = vo*cos(theta)*t is also mentioned, as there is no acceleration in the horizontal direction. The conversation then moves on to finding the initial velocity of an object shot at 70 degrees and taking 6.5s to land. The equation Range = v02sin2θ/g is used, but there is uncertainty if it can be applied in this scenario. Finally, the equation y = vo
://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...

In the horizontal direction there is no acceleration.

s = vo*cos(theta)*t.

rl.bhat said:
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

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.

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.

## 1. How do I calculate the time traveled from launch to land?

To calculate the time traveled from launch to land, you will need to know the distance traveled and the speed of the object. Divide the distance by the speed to get the time traveled. Remember to use the same units for distance and speed.

## 2. Can I use the average speed to calculate the time traveled from launch to land?

Yes, you can use the average speed to calculate the time traveled from launch to land. However, if the speed varies during the journey, it is best to use the equation for average speed which is (total distance traveled)/(total time taken).

## 3. What units should I use for calculating time traveled from launch to land?

It is important to use consistent units when calculating time traveled from launch to land. The most commonly used units are seconds, minutes, and hours. Make sure to use the same units for both distance and speed.

## 4. What factors can affect the accuracy of the calculated time traveled from launch to land?

The accuracy of the calculated time traveled from launch to land can be affected by factors such as air resistance, changes in speed or direction, and external forces such as gravity. It is important to consider these factors when making calculations.

## 5. Are there any other equations or methods for calculating time traveled from launch to land?

Yes, there are other equations and methods for calculating time traveled from launch to land. One example is using the equation for displacement, velocity, and acceleration (d = vt + 1/2at^2). Another method is using a graph and calculating the area under the curve to determine the time traveled. However, the basic equation of distance divided by speed is the most commonly used method.

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