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

Jun 2017
337
6
Lima, Peru
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!
 
Last edited:

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
What/where is point A ?
 
Jun 2017
337
6
Lima, Peru
What/where is point A ?
I'm sorry I forgot to add that detail to the question. Now the picture reflects the way how this problem was intended. Can it be solved?
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
A relatively easy method is to choose a nice value for $v_{oy}$ (20 m/s is a very "nice" choice) and compute the max height and total time the projectile is airborne. From there, calculate the time required to achieve half the max height the second time it passes through that height which, because of symmetry, will be the same time period asked by the problem. Finally, divide that time by the total time to get the desired ratio.

I have worked out another method that involves area under a velocity vs time graph and similar triangles that uses only variables.

Why don't you attempt the easier method and post up your solution.
 
Last edited:
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Jun 2017
337
6
Lima, Peru
A relatively easy method is to choose a nice value for $v_{oy}$ (20 m/s is a very "nice" choice) and compute the max height and total time the projectile is airborne. From there, calculate the time required to achieve half the max height the second time it passes through that height which, because of symmetry, will be the same time period asked by the problem. Finally, divide that time by the total time to get the desired ratio.

I have worked out another method that involves area under a velocity vs time graph and similar triangles that uses only variables.

Why don't you attempt the easier method and post up your solution.
Wait a sec. How did you obtained a "nice choice" of $v_{oy}=20\,frac{m}{s}$. Actually I was looking for some sort of algebraic justification. I've attempted to do that but I came stuck. The other method which you attempted can you post it?.
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
The time a projectile is airborne in a gravitational field, with no forces acting in the horizontal direction, is solely dependent on motion in the vertical direction.

Using the kinematics equation in the vertical direction, $v_{yf}^2 = v_{y0}^2 - 2g\Delta y$

from launch to max vertical height $H$ ...

$ 0^2 = v_{y0}^2 - 2gH \implies H = \dfrac{v_{y0}^2}{2g}$

Let $v_y$ be the vertical velocity at height $\dfrac{H}{2}$

from height $\dfrac{H}{2}$ to max height $H$ ...

$0^2 = v_y^2 - 2g \left(\dfrac{H}{2} \right) \implies 0 = v_y^2 - gH = v_y^2 - g\left(\dfrac{v_{y0}^2}{2g}\right) \implies v_y^2 = \dfrac{v_{y0}^2}{2} \implies v_y = \dfrac{v_{y0}}{\sqrt{2}}$



using the kinematics equation $\Delta y = \dfrac{1}{2}\left(v_0 + v_f \right) \cdot t$

Let $t$ = time from launch to landing as expressed in the original problem statement and $t_1$ = the time elapsed from height $\dfrac{H}{2}$ to max height $H$ ...

$H = \dfrac{1}{2}\left(v_{y0} + 0\right) \cdot \dfrac{t}{2}$
-----------------------------------------
$\dfrac{H}{2} = \dfrac{1}{2}\left(\dfrac{v_{y0}}{\sqrt{2}} + 0\right) \cdot t_1$

dividing the above two equations yields ...

$2 = \dfrac{t}{\sqrt{2} \cdot t_1} \implies t_1 = \dfrac{t}{2\sqrt{2}}$

The time elapsed from initially reaching height $\dfrac{H}{2}$ to landing is $\dfrac{t}{2} + t_1 = t\left(\dfrac{1+\sqrt{2}}{2\sqrt{2}}\right) \approx 0.85t$

Wait a sec. How did you obtained a "nice choice" of $v_{oy}=20 \frac{m}{s}$
Using $v_{y0} = 20 \text{ m/s } \implies$ it takes 2 seconds to reach max height $\implies t_{total} = 4 \text{ sec}$

because of symmetry, time from launch to the 2nd time the projectile reaches height $\dfrac{H}{2}$ = time from initially at height $\dfrac{H}{2}$ to landing on the ground.

$H = \dfrac{v_{y0}^2}{2g} = \dfrac{20^2}{20} = 20 \text{ m}$

$\Delta y = v_{y0} \cdot t - \dfrac{1}{2}gt^2$

$10 = 20t - 5t^2 \implies t^2 - 4t + 2 = 0 \implies t^2 - 4t + 4 = 2 \implies (t-2)^2 = 2 \implies t = \sqrt{2} \pm 2$

ratio of the later time to the total time = $\dfrac{\sqrt{2}+2}{4} \approx 0.85$
 
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