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For a missile travelling from (0,0) at angle $\theta$ (to the horizontal) and initial velocity $u$, the y (vertical) position at time t is given by

$s_{y} = u\sin (\theta) t - 0.5gt^{2}$

and the x position id

$s_{x} = u\cos(\theta)t$,

but I am struggling to calculate the angle $\theta_{t}$ at time $t$ to the horizontal. For example when the missile reaches its apex the angle is zero, after which it will be negative and hopefully on arrival it will be $-\theta$.

So my thinking is to differentiate $s_{y}$ w.r.t. $t$ I will in effect get the gradient and hence I can arctan this to get the angle $\theta_{t}$.

$s'_{y} = u\sin (\theta) - gt$


$\theta_{t} = \arctan(u\sin (\theta) - gt)$

Is this line of thinking correct? If not, lease explain how to figure this out. Thanks in advance.

EDIT I should admit that I am sure what I have done is not correct as it doesn't even work for $t=0$, but I am at a loss as to how to figure this (seemingly easy) problem out.

EDIT 2 Another idea is to simply do

$gradient = s_{y}(t)/S_{x}(t)$ at time $t$ and then $\theta_{t} = \arctan(\frac{u\sin (\theta) t - 0.5gt^{2}}{u\cos(\theta)t})$ which works much better.

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I think you could use $tan(\theta) = \frac{y}{x}$ where $y$ the velocity in $y$ direction and $x$ is velocity in $x$ direction at the same time – Dimensionless Mar 24 '13 at 15:04
up vote 1 down vote accepted

The error in your reasoning is to use the equations that hold at launch of the projectile, t=0, and apply them at any time later.

Once you write




these equations hold for any subsequent time $t$.

The angle $\theta(t)$ that you are looking for, during flight, is the angle of the tangent to the path of the projectile. So you cane write down

$\tan(\theta(t))=\frac{u\sin(\theta_0)-gt}{u\cos(\theta_0)} $

from which you get

$\theta(t)=\tan^{-1}\left( \frac{u\sin(\theta_0)-gt}{u\cos(\theta_0)}\right) $

Check to see that this equation gives the angle $\theta(t)$ correctly provided you use values of $t$ in the interval [0,$t_f$] where $t_f$ is the total time of flight.

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