# How to find the height at which a projectile is launched from an angle? [closed]

I need guidance with the following question:

A boat is $50.0m$ from the base of a cliff, fleeing at $5.00m/s$. A gun mounted on the edge of the cliff fires a shell at $40.0m/s$ and hits the boat when it has fled another $50.0m$ What is the height of the cliff?

Here is what I have done so far (Note: The values used were calculated from a previous part of my question):

Additional info: The gun is fired at an angle of $18.9^\circ$ from the horizontal up and the trajectory has a parabolic shape.

I found the angle by using $\Theta=\frac {1}{2} sin^-$ $^1$ $(\frac {-a\Delta d_x}{v_1^2})$ which obtained from substituting and rearranging equations.

$\Delta d_x=v_1$$_x$$\Delta t+\frac {1}{2}a\Delta t^2$

$\Delta d_x=v_1$$_x$$\Delta t$

$\Delta t=\frac {\Delta d_x}{v_x}$

Substitute $\Delta t$ of the above equation into the equation below.

$\Delta d_y=v_1$$_y$$\Delta t+\frac {1}{2}a\Delta t^2$

$\Delta d_y=v_1$$_y$$\frac {\Delta d_x}{v_x}+\frac {1}{2}a(\frac {\Delta d_x}{v_x})^2$

Now this here below is where I am unsure if I am supposed to use the initial vertical velocity I calculated from a previous part of the question. I believe I am not supposed to use it and that it is irrelevant to the problem since I am trying to find the height of the cliff ($\Delta d_y$) where there is no initial vertical velocity. However I am not sure of this and would like someone more knowledgeable in projectile motion to tell me if what I said is right or not and what it should be.

$\Delta d_y=(13.0m/s)(\frac {100.0m}{37.6m/s})+\frac {1}{2}(-9.8m/s^2)(\frac {100.0m}{37.6m/s})^2$

This is not a homework question, it is an exercise for an upcoming test. I need to know for these types of projectile problem if I need to use the initial vertical velocity.

## closed as off-topic by John Rennie, ACuriousMind♦, user36790, Kyle Kanos, MartinOct 13 '15 at 11:50

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• Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie Oct 13 '15 at 6:23
• @JohnRennie Sorry this is not homework its an exercise question for an upcoming test. I am just basically asking if I am supposed to use initial vertical velocity or not. That is all. – Scienceguy Oct 13 '15 at 6:25
• @Scienceguy What is the direction in which the gun fires the shell? Downwards or horizontal or....? – SchrodingersCat Oct 13 '15 at 6:42
• @Aniket Well the gun fires the shell 18.9 degrees above the horizontal and the trajectory has a parabolic shape. – Scienceguy Oct 13 '15 at 6:47
• Hi Scienceguy. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Oct 13 '15 at 7:53

$\Delta d_y=v_1$$_y$$\frac {\Delta d_x}{v_x}+\frac {1}{2}a(\frac {\Delta d_x}{v_x})^2$

$\Delta d_y=(13.7m/s)(\frac {100.0m}{37.6m/s})+\frac {1}{2}(-9.8m/s^2)(\frac {100.0m}{37.6m/s})^2$

The question of whether you should put a $v_1$$_y$ really depends on your choice of initial condition. Take note that the cannonball starts at cliff height (1), flies higher than that to a peak height (2), then falls back to cliff height (3), and finally goes to sea level (4) and hits the boat.

Time is 0 at (1) while, at (2) and (3) these are easy to calculate since you have the initial velocity of the cannonball as well as it's launch angle giving you vertical launch velocity, and you also have g. I would also point out that 13.7m/s seems like a wrong value to use anyway. Vertical lauch velocity is $40 \sin 18.9$ (=12.96) and horizontal launch velocity is $40 \cos 18.9$.

Now there are exactly only two points in time where the ball has a vertical velocity of 12.96, that is at time (1) and time (3). Very importantly, it is +12.96 at (1) and -12.96 at (3), as I note you have chosen acceleration due to gravity to be negative 9.8. (ball is going up at t=0, and ball is going down as it passes the cliff height again). I cannot stress more the importance of the signs here, since the initial condition chosen allows for the ball to have travelled in both directions in the y axis with the cannon at y=0.

Now if you wish to plug in +12.96 as $\Delta d_y$ into the final equation, then t must be 10, not $\frac {100.0m}{37.6m/s}$, because +12.96 is the vertical velocity of the cannonball at launch time, and the ball only spends 10 seconds in the air before impact (50m travelled by the 5m/s boat during the ball's full, parabolic airtime)

You get your answer as a negative value, more specifically it is the height of the cliff relative to the cannon, -360m.

If you don't want to deal with initial vertical velocity, your only chance is to take (2) as the initial condition, where the ball is the highest it will ever go. in which case the t you use is 10 minus some small amount it takes for the ball to reach that height in the first place. I recommend you verify this result for yourself and also see if you can achieve the same answer using position (3) as the initial condition.

It's not clear to me how you have got the values of $v_x$ and $v_y$ you used in the final part of the question and given the info in the question I don't think they are correct.

I think you are missing a critical point at the start.

You can easily calculate the time of flight as 10s as you know the boat speed and how far it traveled in that time.

You can then put this value into your first set of equations to get the $v_x$ of the shell.

As you know the initial total $v$ of the shell you can then calculate the angle it was fired at and hence $v_y$ (in fact you don't even need to calculate the angle).

Using $v_y$ and $t$ you can calculate $d$.

• The values of $v_x$ and $v_y$ don't really matter with what I am asking. I simply want to know if I am supposed to use the initial velocity ($13.7m/s$) in my final calculation (the last line) to find the height of the cliff ($\Delta d_y$). – Scienceguy Oct 13 '15 at 10:36
• Yes you need to use the initial vertical velocity to calculate the height of the cliff and you have the right equation for that, but I think you have the wrong value for the velocity, unless there is more info in the question that you haven't said. – nivag Oct 13 '15 at 10:41
• Yes, sorry the $v_y$ is wrong, it is supposed to be $13.0m/s$. The angle from which the gun is fired is $18.9^\circ$ degrees which I used with the initial velocity of the shell to find the $v_y$. $40.0m/ssin(18.9^\circ)=13.0m/s$ – Scienceguy Oct 13 '15 at 10:48
• But how do you get the angle, it is not in the question? – nivag Oct 13 '15 at 10:50
• I calculated it from a previous part of the question. I have now included it in the post. I came up with the equation $\Theta=\frac {1}{2} sin^-$ $^1$ $(\frac {-a\Delta d_x}{v_1^2})$ – Scienceguy Oct 13 '15 at 10:53