Question

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 if air resistance is negligible.

Short answer

The time required by a projectile to reach its highest point is $t=\frac{v_0\sin \theta }{g}$. This time is equal to the time required by the projectile to return to its original height.

Step-by-step solution

Step 1. Kinematic equations for projectile motion

A projectile that is projected from a height yabove the ground with initial velocity${\overrightarrow{v}}_0$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:

• For the horizontal motion of projectile:

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

• For the vertical motion of projectile:

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

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.

Step 2. Figure showing the projectile motion

Consider that a projectile is projected from height ywith the initial velocity $v_0$at an angle $\theta$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.,$v{\text{'}}_{\mathrm{y}}=0\mathrm{m}/\mathrm{s}$.

Step 3. Determination of the time taken by the projectile to reach its maximum height

Let $v_{\mathrm{x}0}$and $v_{\mathrm{y}0}$be the horizontal and vertical components of the initial velocity ${\overrightarrow{v}}_0$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 ${\overrightarrow{v}}_0$of the projectile in terms of $\theta$as:

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

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

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

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

Step 4. Determination of the maximum height reached by the projectile above the reference level

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

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

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

Step 5. Determination of the time required by the projectile to return to its original height

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:

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

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

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.