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A projectile that is projected from a height yabove the ground with initial velocity${\stackrel{鈫�}{v}}_0$follows a parabolic path. Suppose the point of projection is taken as the origin and the upward direction as the positive y-axis. In that case, the kinematics equations for the projectile motion can be written as:

$\begin{array}{c}{v}_{\mathrm{x}}=v_{\mathrm{x}0}\\ x=v_{\mathrm{x}0}t\end{array}$

$\begin{array}{c}{v}_{\mathrm{y}}=v_{\mathrm{y}0}-gt\\ y=v_{\mathrm{y}0}t-\frac{1}{2}g{t}^2\\ v_{\mathrm{y}}^2=v_{\mathrm{y}0}^2-2gy\end{array}$

Here, xandy are the horizontal and vertical displacements of the projectile traveled in time t, $v_{\mathrm{x}0}$ and$v_{\mathrm{y}0}$ are the horizontal and vertical components of the initial velocity of the particle, and$v_{\mathrm{x}},v_{\mathrm{y}}$ are the horizontal and vertical components of velocity of the particle at time t.

Consider that a projectile is projected from height ywith the initial velocity $v_0$at an angle $胃$as shown in the figure below. This projectile follows a parabolic path, as shown in the figure below:

Take the point of launching the projectile, i.e., point O as the origin and the upward direction to be the positive y-direction. Therefore, the acceleration of the projectile will be equal to the negative of the acceleration due to gravity, i.e.,$a=-g$.

The projectile reaches its maximum height at point A. Let the height of this point above the reference level be $y\text{'}$. At point A, the velocity of the particle will be along the horizontal; therefore, the y component of velocity will be zero, i.e.,$v{\text{'}}_{\mathrm{y}}=0\mathrm{m}/\mathrm{s}$.

Let $v_{\mathrm{x}0}$and $v_{\mathrm{y}0}$be the horizontal and vertical components of the initial velocity ${\stackrel{鈫�}{v}}_0$of the projectile, and t be the time taken to reach the maximum height at point A.

You can write the values of the horizontal and vertical components of the initial velocity ${\stackrel{鈫�}{v}}_0$of the projectile in terms of $胃$as:

$\begin{array}{c}{v}_{\mathrm{x}0}=v_0\cos 胃\\ v_{\mathrm{y}0}=v_0\sin 胃\end{array}$

Use the kinematic equation for the vertical motion of the projectile from points O to A as:

role="math" localid="1644472631072" $\begin{array}{c}{v}_{\mathrm{y}}=v_{\mathrm{y}0}-gt\\ 0=v_0\sin 胃-gt\\ t=\frac{v_0\sin 胃}{g}...\left(i\right)\end{array}$

This is the time required by the projectile to reach its highest point.

Using the kinematic equation for the vertical motion of the projectile at this point, you will get:

$\begin{array}{c}{\left({v\text{'}}_{\mathrm{y}}\right)}^2=v_{\mathrm{y}0}^2-2gy\text{'}\\ y\text{'}=\frac{v_{\mathrm{y}0}^2-{\left({v\text{'}}_{\mathrm{y}}\right)}^2}{2g}\\ =\frac{{\left(v_0\sin 胃\right)}^2-{\left(0\right)}^2}{2g}\end{array}$

role="math" localid="1644472772674" ${\left({v\text{'}}_{\mathrm{y}}\right)}^2=\frac{v_0^2{\sin }^2胃}{2g}...\left(ii\right)$

The projectile returns from the point of maximum height A to its original height at point B. This is shown in the figure above.

Now, during the motion of the projectile from points A to B, the initial velocity of the particle at point A will be $v{\text{'}}_{\mathrm{y}}=0\mathrm{m}/\mathrm{s}$and the final velocity of the particle at point B will be ${v_{\rm{y}}}$. Since the motion of the particle is downward, so $y\text{'}$is taken as negative.

Let the time taken by the particle to return to its original height from the point of maximum height be $t\text{'}$.

Using the kinematic equation for the vertical motion of the projectile from points A to B, you will get:

$\begin{array}{c}-y\text{'}={v\text{'}}_{\mathrm{y}}t-\frac{1}{2}g{\left(t\text{'}\right)}^2\\ -y\text{'}=\left(0\right)t-\frac{1}{2}g{\left(t\text{'}\right)}^2\\ {\left(t\text{'}\right)}^2=\frac{2y\text{'}}{g}\end{array}$

On substituting the value of $y\text{'}$from equation (ii) in the above expression, you will get:

${\left(t\text{'}\right)}^2=\frac{2}{g}\left(\frac{v_0^2{\sin }^2胃}{2g}\right)$

role="math" localid="1644472904332" $t\text{'}=\frac{v_0\sin 胃}{g}...\left(iii\right)$

From (i) and (iii), it is clear that:

$t\text{'}=t$

Thus, you can say that the time required by a projectile to reach its highest point is equal to the time required to return to its original height.