# Homework Help: Pre-Lab derivation

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1. Oct 13, 2015

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.

3. 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

Last edited by a moderator: Oct 13, 2015
2. Oct 14, 2015

### haruspex

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. Oct 14, 2015

### andrevdh

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. Oct 14, 2015

### haruspex

If both v's are velocities the equation is dimensionally wrong. And g should feature somewhere.

5. Oct 14, 2015

### andrevdh

I am referring to the equation in 1.

6. Oct 14, 2015

### haruspex

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. Oct 14, 2015

### andrevdh

Sorry. Will use it in the future.

8. Oct 14, 2015

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. Oct 14, 2015

### Geofleur

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. Oct 14, 2015

### haruspex

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. Oct 15, 2015

### andrevdh

What will the y-displacement be when the projectile lands back on the ground?

Last edited: Oct 15, 2015