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?. Can someone help me here?. Please!

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?. Can someone help me here?. Please!

Last edited: