# Horizontal projectile motion, finding the height of the object at a given time. [closed]

A canon is fired Horizontal to the ground 80 meters above ground The canon ball is fired a $T=0$ and hits the ground at $T_g$. Calculate the height at the time $T_g/2$ So far I have calculated time. which gives me the equation $$y=y_i+v_{iy}(4.05) - (0.5)(9.82)(4.05^2)$$ Now my problem is that when I try to calculate $v_{iy}$ I am getting $-37\text{ m/s}$ which I know is not the case; it wouldn't make any sense. $Y$ initial is equal to $80\ \ V_{iy}$ I don't know $4.05$ is the time it takes the projectile to hit the ground. $9.82$ is gravity which is my acceleration.
So my question is two parts. First, is this the best equation to find what I'm trying to find? Second, how do I find velocity in the $y$ direction when $\theta$ is $180$ degrees? This is an edited question I do apologize for being unclear. Thank you for responding and requesting clarification. I was using the incorrect equation a simpler way to find the answer would be $y=.75\times H$. Because change in $y = -.5\times g\times t^2$ sense initial velocity in the $y$ direction equals $0$. Thank you for the help in structuring my question better.

• Your question does not make sense as is. There isn't any diagram or context description, only a formula with numbers that popped out of nowhere, and the question that follows is unclear. Please edit your post with a clear description of the problem, what you have done so far in algebra then the numerical application, and a clear question. We can still guess and I'm going to answer but you should really pay attention to that or you'll have very little chance of having answers in the future. Mar 2 '14 at 0:51
• I have updated what I have done so far I am not sure if this is what you wanted is there a chance you could point me in the direction of what you would consider a better asked questions so I can in the future ask a better question? If not could you tell me what else I am missing? Mar 2 '14 at 1:03
• Hello Mr./Ms. Moderator, I am unsure as to why this has been flagged, is there anyway I could be told more detail than the standard message about something being off topic Also as I wanted to be guided wasn't looking for an answer but an equation I do not see how I am breaking any of your stated rules. At this point I have asked several question all of which have been answered to my satisfaction even though one was edited by me and another was edited by someone else, I am afraid all of my questions break this unknown rule. Mar 2 '14 at 5:59
• In addition I would like to refute each point in specific so if anyone does respond they can show me exactly wher eI am Misunderstanding the rules: Ask about a specific physics concept: I asked about horizontal projectile equation I figure thats fairly specific. Show some effort to work through the problem: I showed what I had done so far had already solved for one variable and explained my understanding about the remaining variable and what I had gotten so far for it. Mar 2 '14 at 6:34
• Useful to the broader community: Considering this is on both my midterm and final as a concept as well as being on many questions with in my homework I think I can safely assume any answer would be useful to someone else at my level of understanding of physics, unless of course I asked a question too far below the intelligence of the community in which case i apologize for asking questions here I will gladly move to another forum that allows freshman college level questions. Mar 2 '14 at 6:35

[Edit: If you don't consider air friction and if you're not asked anything about the range of the projectile, the equations are the same as a vertical freefall]. If you are trying to find the velocity of the object at any given time, it is not $v_{iy}$ that you need to calculate since it is the initial velocity of the object at $t=0$.
Step by step for a vertical freefall (1D) with origin on the ground at the vertical of the initial position of the object and y axis toward the object: $$-g=a_y$$ $$\Rightarrow v(t) = \int^t_0 -g.dt=-g.t+v_{0y}$$ $$\Rightarrow y(t) = \int^t_0 (-g.t+v_{0y}).dt=-\frac{1}{2}g.t^2+v_{0y}.t+y_0$$ Here $v_{0y}$ is 0 and $y_0=80m$. You're interested in $v(T_g/2)$ where $T_g$ is $t$ so that $y(t)=0$.
• If you don't have any questions about the range of the projectile or anything, you only need the y axis. This is considering that there isn't any air friction, of course, and that is why you haven't been given the initial horizontal velocity. Plug in $y_0$ in y(t) to solve for t, which you have done, then plug that t (named $T_g$ apparently) in v(t). Done. Mar 2 '14 at 1:17