Calculate time traveled from launch to land

  • Thread starter Thread starter ://Justice
  • Start date Start date
  • Tags Tags
    Launch Time
AI Thread Summary
To calculate the time from launch to landing with a velocity of 30.197 m/s and a distance of 85 m, the initial approach using the formula X = Xo + Vot + 0.5at^2 was initially confusing. The user realized that a simpler method, using Distance/Velocity = Time, yielded a time of 2.815 seconds, which seemed short compared to the empirical data of 4 seconds. After further consideration, they used the horizontal motion equation s = vo*cos(theta)*t, resulting in a more reasonable time of 3.356 seconds. The discussion then shifted to finding the initial velocity for a projectile launched at 70 degrees that took 6.5 seconds to land, emphasizing the importance of analyzing vertical components. The suggested approach involved using the vertical displacement equation to solve for the initial velocity.
://Justice
Messages
27
Reaction score
0

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...
 
Physics news on Phys.org
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top