projectile & time

[logistic_guy]

Show that the time required for a projectile to reach its highest point is equal to the time for it to return to its original height.

 

[khansaheb]

Show that the time required for a projectile to reach its highest point is equal to the time for it to return to its original height.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

First equation can give us the time.

$\displaystyle v = v_0 + at = v_0 - gt$

At the highest point $\displaystyle v = 0$, then

$\displaystyle 0 = v_0 - gt$

This gives:

$\displaystyle t = \textcolor{blue}{\frac{v_0}{g}} \ \ \$ (time to go to the highest point.)

Second equation can give us the height.

$\displaystyle v^2 = v^2_0 + 2ah = v^2_0 - 2gh$

At the highest point $\displaystyle v = 0$, then

$\displaystyle 0 = v^2_0 - 2gh$

This gives:

$\displaystyle h = \frac{v^2_0}{2g}$

Third equation can give us the time to go to the lowest point.

Suppose now the projectile is at the highest point and it is gonna go to the lowest point.

$\displaystyle y = y_0 + v_0t + \frac{1}{2}at^2 = y_0 + v_0t - \frac{1}{2}gt^2$

At the highest point, the initial velocity $\displaystyle v_0 = 0$ and the initial height $\displaystyle y_0 = h$, then

$\displaystyle 0 = h + (0)t - \frac{1}{2}gt^2$

$\displaystyle 0 = h - \frac{1}{2}gt^2$

Solving for the time gives:

$\displaystyle t = \sqrt{\frac{2h}{g}}$

But we already know the height from the second equation, then

$\displaystyle t = \sqrt{\frac{2\frac{v^2_0}{2g}}{g}} = \sqrt{\frac{v^2_0}{g^2}} = \textcolor{red}{\frac{v_0}{g}} \ \ \$ (time to go to the lowest point.)

This shows that:

time to go to the highest point $\displaystyle =$ time to go to the lowest point

Or

$\displaystyle \textcolor{blue}{t_{\text{up}}} = \textcolor{blue}{\frac{v_0}{g}} = \textcolor{red}{\frac{v_0}{g}} = \textcolor{red}{t_{\text{down}}}$