Deriving the Equation for Projectile Motion: t = 2(vi)/g*sin(theta)

In summary, the conversation revolves around deriving the equation t = 2(vi)/g*sin(theta) which represents the time for a projectile launched at an angle θ to reach the ground level again. The context is a launch simulation and the given equation is to be derived from first principles. Some confusion arises regarding the placement of sin(theta) in the equation, as well as the use of v and g. The hint is given to use the equation for the vertical component of velocity under constant acceleration to derive the total time of flight.
  • #1
Paradise Jack
2
0
1. Derive this equation
t = 2(vi)/g*sin(theta)

where vi isinitial velocity, and g is acceleration due to gravity

2. Implicit differentiation, possibly gravity as a constant.

The Attempt at a Solution



t = 2(vi)/v*sin(theta)
dt/d(theta) = 2(vi)/g * dt/d(theta)(sin(theta))
dt/d(theta) = 2(vi)/g * cos(theta)

Basically, I said that velocity and gravity were constants as well as 2. Would this be a correct derivation of the listed formula
 
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  • #2
First, a physical equation is meaningless without a context and a definition of variables within that context. Describe the set-up.
Next, I don't understand the first equation of your attempted solution. What is v here, and where does this equation come from?
 
  • #3
I think this is the formula for the time for a projectile launched from ground level at an angle θ to reach the ground level again, in which case the sin(θ) term should be in the numerator not the denominator.
 
  • #4
andrevdh said:
I think this is the formula for the time for a projectile launched from ground level at an angle θ to reach the ground level again, in which case the sin(θ) term should be in the numerator not the denominator.
If both v's are velocities the equation is dimensionally wrong. And g should feature somewhere.
 
  • #5
I am referring to the equation in 1.
 
  • #6
andrevdh said:
I am referring to the equation in 1.
Ok. It does help to use the quote feature or otherwise clarify the reference. From the sequence of posts, it was natural to read it as a response to my comment on an equation.
To reply appropriately to your post, there being no parentheses, the sin is in the numerator.
 
  • #7
Sorry. Will use it in the future.
 
  • #8
Alright, I'm going to update this for you since apparently there is a lot of confusion. g is that gravitational constant of acceleration. V(i) is the initial velocity, as stated above. The context is a launch simulation, and the problem asks for a derivation. The lack of parentheses does say that the sin(theta) is in the numerator, and it is absolutely correct. Whether it should be there or not in your opinion, it is there in my homework.
 
  • #9
It seems you are to derive the time a projectile spends in the air (or vacuum, since we generally neglect air resistance for this problem) between launch and touch-down. Here is a hint: The vertical part of the motion is subject to constant acceleration, and there is an equation for the vertical component of velocity in that case. At the top of the trajectory, the vertical component of velocity is zero (why?). This fact will enable you to derive the time it takes for the projectile to reach the highest point of the trajectory. The total time is twice this time (why?).
 
  • #10
Paradise Jack said:
Alright, I'm going to update this for you since apparently there is a lot of confusion. g is that gravitational constant of acceleration. V(i) is the initial velocity, as stated above. The context is a launch simulation, and the problem asks for a derivation. The lack of parentheses does say that the sin(theta) is in the numerator, and it is absolutely correct. Whether it should be there or not in your opinion, it is there in my homework.
Ok, and I assume the 'v' (as opposed to v(i)) in the first equation of your attempted solution is a typo for 'g'.
If so, I wonder if you have misunderstood what you are asked to do. You are asked to show that the given equation is correct (derive it, i.e. obtain it from first principles). Instead, you seem to have taken it as correct and differentiated it.
 
  • #11
What will the y-displacement be when the projectile lands back on the ground?
 
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FAQ: Deriving the Equation for Projectile Motion: t = 2(vi)/g*sin(theta)

1. What is a pre-lab derivation?

A pre-lab derivation is a written explanation of the mathematical equations and principles used in an experiment before it is conducted. It outlines the theoretical background and reasoning behind the experiment and helps the scientist understand the expected results.

2. Why is a pre-lab derivation important?

A pre-lab derivation is important because it allows the scientist to fully understand the experiment and its expected outcomes. It also serves as a reference for the scientist during the experiment and helps ensure accuracy and validity of the results.

3. How do you write a pre-lab derivation?

To write a pre-lab derivation, the scientist should first review the experimental procedure and identify the key equations and principles involved. They should then explain the derivation of these equations, including any assumptions made. The derivation should be written in a clear and concise manner, with all steps and calculations clearly shown.

4. What should be included in a pre-lab derivation?

A pre-lab derivation should include a brief introduction to the experiment, a description of the relevant equations and principles, and a step-by-step derivation of these equations. It should also include any assumptions made and a discussion of the expected results based on the derived equations.

5. Are pre-lab derivations necessary for all experiments?

No, pre-lab derivations are not necessary for all experiments. They are typically used in more complex experiments that involve mathematical equations and principles. However, it is always beneficial for the scientist to have a thorough understanding of the experiment and its underlying concepts, even if a formal derivation is not required.

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