How do I find the time elapsed from a given point in a parabollic trajectory until the impact to the ground?

Question

The problem is as follows:

The diagram from below shows a projectile being fired from the origin of coordinates. It is known that it takes $t$ seconds to impact the ground at a given range $R$. Find the time that it takes from point $A$ to the point of impact in $R$. Assume $g=10\,\frac{m}{s^2}$.

The alternatives given are as follows:

$\begin{array}{ll} 1.&0.60t\\ 2.&0.65t\\ 3.&0.70t\\ 4.&0.85t\\ \end{array}$

I'm confused exactly on how to manipulate the equations for a parabollic trajectory.

In this given situation it can be established that the vertex is at $H,\frac{R}{2}$.

Thus the equation for the parabola would be as follows:

$y=-a\left(x-\frac{R}{2}\right)^2+H$

Since the points in the trajectory which are known are $(0,0)$ and $(R,0)$ it can be known the value of a:

$0=-a\left(0-\frac{R}{2}\right)^2+H$

$a=\frac{4H}{R^2}$

Then:

$y=-\frac{4H}{R^2}\left(x-\frac{R}{2}\right)^2+H$

Therefore when the $\frac{H}{2}$ then $x$ would be:

$\frac{H}{2}=-\frac{4H}{R^2}\left(x-\frac{R}{2}\right)^2+H$

$x=-\frac{1}{4}\left(-2 + \sqrt 2\right) R$

$x=\frac{1}{4}\left(2 + \sqrt 2\right) R$

Since in the graph it is observed that the range is in the first quadrant then I'm discarding the negative value, and it becomes reduced to:

$x=\frac{1}{4}\left(2 + \sqrt 2\right) R$

Which should be value of $R$.

But then this is where I'm stuck, how am I supposed to get the time elapsed from this equation?. Can somebody help me here?.

The only equations which I recall are:

$y=v_o\sin\omega t -\frac{1}{2}gt^2$

$x=v_o \cos\omega t$

But that's where I'm still stuck. Can someone indicate me what sort of algebraic manipulation should I do to obtain the requested time?.

Answer 1

In this case you can split the motion. You can see the problem as the projectile drop from H to 0 and drop from H to H/2.

So the t equals to sqrt(2H/g)+sqrt(H/g)

Answer 2

Hint:

The motion is described by the equations: $$ \begin{cases} y=-\frac{1}{2}gt^2+v_yt\\ x=v_xt \end{cases} $$ Use the known points: $(0,0), (R,0), (R/2,H)$ to find $v_x$ and $v_y$, then you can find the time needed to go tot the point $A$... and solve the problem.

Answer 3

*I renamed the time to reach the point $R$ as $tf$, to avoid confusion with the time variable $t$.

The velocities along the two axes are: \begin{cases} v_y = -gt + v_y(0)t\\ v_x = constant = \frac{R}{t_f}. \end{cases}
The motion is described by the following equations: \begin{cases} y(t)=-\frac{1}{2}gt^2+v_y(0)t\\ x(t)=v_xt \end{cases}
Describing a point as: $P(t,y)$, the point A corresponds to: $A(t_a, \frac{H}{2})$.
Note that there will be two different points with height $\frac{H}{2}$, but A is the time with the smaller time value.

Focusing on the $y(t)=-\frac{1}{2}gt^2+v_y(0)t$,
we know that $y(t)$ intersect the poins $(0,0), (H,\frac{t_f}{2}), (0, t_f).$

Using the third listed above we find the equation:
$0 = \frac{g}{2}t_f^2 + v_y(0)t_f$.

Solving for $t_f$ we find:

$ t_f=\frac{2v_y(0)}{g}, v_y(0)=\frac{gt_f}{2} $.

We rewrite $y(t)$:
$y(t)=-\frac{g}{2}t^2 + \frac{gt_f}{2}t = \frac{gt}{2}(t_f-t).$

Now we want to find the time $t_a$ where $t_a$ is the time where (for the first time):
$y(t)=\frac{H}{2}$.

Using the point $(H,\frac{t_f}{2})$ we find $H$ to be:
$H=\frac{gt_f^2}{8}$.

Solving the system:
\begin{cases} y(\frac{H}{2})=\frac{gt}{2}(t_f-t)\\ H=\frac{gt_f^2}{8} \end{cases} we find $t_a$ to be equals to $0,15t$.

So finally we determine the time to go from point A to point R, $t_{AR}$:
$t_{AR}=t_f-t_a=t_f-0.15t_f=0.85t_f$.