Question

Revisit Example 3–7, and assume that the boy with the slingshot is below the boy in the tree (Fig. 3–40) and so aims upward, directly at the boy in the tree. Show that again the boy in the tree makes the wrong move by letting go at the moment the water balloon is shot.

FIGURE 3-40Problem 36

Short answer

The second boy in the tree makes the wrong move by falling from the tree at the same time the water balloon is shot.

This is because the boy and the water balloon take the same time to reach a certain height above the ground, and therefore the water balloon hits the boy at this height.

Step-by-step solution

Step 1. Kinematic equations for projectile motion

If any object projected from a height yabove the ground with initial velocity${\overrightarrow{v}}_0$follows the parabolic path, it is termed projectile. If the upward direction is taken to be positive, then the kinematics equations for projectile motion can be written as:

• For the horizontal motion of projectile:

$\begin{array}{c}{v}_{\mathrm{x}}=v_{\mathrm{x}0}\\ x=x_0+v_{\mathrm{x}0}t\end{array}$

• For the vertical motion of projectile:

$\begin{array}{c}{v}_{\mathrm{y}}=v_{\mathrm{y}0}-gt\\ y=y_0+v_{\mathrm{y}0}t-\frac{1}{2}g{t}^2\\ v_{\mathrm{y}}^2=v_{\mathrm{y}0}^2-2gy\end{array}$

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 the velocity of the particle at time t.

Step 2. Given information

The initial velocity with which the water balloon is shot by the slingshot is $v_0$.

The water balloon leaves the ground at an angle $\theta$.

The horizontal distance between the boy with the slingshot (projectile) and the tree is $x=d$.

If the upward direction is taken to be positive, then the acceleration of the water balloon and boy falling from the tree will be equal to the negative of the acceleration due to gravity, i.e.,$a=-g$

Let the initial vertical height of the boy with the slingshot be $y_1$and the boy hanging on the tree be $y_2$. Consider that the boy and the water balloon collide at height ${y_1}^{\text{'}}$above the ground.

Step 3. Determination of time taken by the water balloon to reach the height at which it hits the boy

Here, our projectile is a water balloon. 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 balloon. Thus,

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

Let t be the time taken by the water balloon to hit the boy.

Using the kinematic equation for the horizontal motion of the water balloon, you can obtain time tas:

$\begin{array}{c}x=v_{\mathrm{x}0}t\\ t=\frac{x}{v_{\mathrm{x}0}}\\ =\frac{d}{v_0\cos {\theta }_0}\end{array}$

Step 4. Determination of the height (or position) of the water balloon above the ground at time t 

Using the kinematic equation for the vertical motion of the water balloon, you will get height ${y_1}^{\text{'}}$as:

$\begin{array}{c}{y\text{'}}_1=y_1+v_{\mathrm{y}0}t-\frac{1}{2}g{\left(t\text{'}\right)}^2\\ {y\text{'}}_1=y_1+\left(v_0\sin {\theta }_0\right)\left(\frac{d}{v_0\cos {\theta }_0}\right)-\frac{1}{2}g{\left(\frac{d}{v_0\cos {\theta }_0}\right)}^2\end{array}$

role="math" localid="1644475053038" $y{\text{'}}_1=y_1+d\tan {\theta }_0-\frac{1}{2}g{\left(\frac{d}{v_0\cos {\theta }_0}\right)}^2...\left(i\right)$

This is the height above the ground where the water balloon reaches at time t.

Step 5. Determination of the position of the boy above the ground at time t

Let $y{\text{'}}_2$be the vertical distance traveled by the boy in time t. The initial velocity of the boy is$u=0\mathrm{m}/\mathrm{s}$.

Using the kinematic equation for free fall of the boy, you will get:

role="math" localid="1644475095012" $\begin{array}{c}{y\text{'}}_2=y_2+ut-\frac{1}{2}g{\left(t\right)}^2\\ =y_2+\left(0\right)t-\frac{1}{2}g{\left(\frac{d}{v_0\cos {\theta }_0}\right)}^2\\ =y_2-\frac{1}{2}g{\left(\frac{d}{v_0\cos {\theta }_0}\right)}^2...\left(ii\right)\end{array}$

Thus, the boy reaches height $y{\text{'}}_2$above the ground at time t.It is clear that the water balloon will hit the boy if $y{\text{'}}_1=y{\text{'}}_2$.

Step 6. Relation between the initial heights of the water balloon and the boy hanging on the tree

It is clear from the figure that in $\Delta \mathrm{OAB}$,

$\tan {\theta }_0=\frac{y_2-y_1}{d}$

role="math" localid="1644475166880" $d\tan {\theta }_0=y_2-y_1...\left(iii\right)$

Step 7. Comparison between the heights of the water balloon and the boy falling from the tree at time t

On substituting (iii) in (i), you will get:

$y{\text{'}}_1=y_1+\left(y_2-y_1\right)-\frac{1}{2}g{\left(\frac{d}{v_0\cos {\theta }_0}\right)}^2$

role="math" localid="1644475226270" $y{\text{'}}_1=y_2-\frac{1}{2}g{\left(\frac{d}{v_0\cos {\theta }_0}\right)}^2...\left(iv\right)$

From (ii) and (iv), it is clear that$y{\text{'}}_1=y{\text{'}}_2$

Thus, youcan say that the boy and the water balloon reach the same height at the same time t, and that the water balloon will definitely hit the boy.

Therefore, the boy hanging on the tree makes the wrong move by falling from the tree at the same time the water balloon is shot.