Question

Why is the following situation impossible? Albert Pujols hits a home run so that the baseball just clears the top row of bleachers,$\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{m}}$high, located$\mathbf{130}\mathbf{}\mathbf{\text{m}}$from home plate. The ball is hit at$\mathbf{41}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{\text{m/s}}$on an angle of${\mathbf{35}}^{\mathbf{\circ }}$to the horizontal, and air resistance is negligible.

Short answer

The given situation is impossible because the calculated vertical distance is$20\text{m}$ which is lesser than the height of the bleaches.

Step-by-step solution

The formula to calculate the vertical distance covered by the baseball.

The velocity with which the ball is hit is $\mathbf{41}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$at an angle of${\mathbf{35}}^{\mathbf{\circ }}$to the horizontal and vertical distance covered is$\mathbf{\text{24m}}$. The horizontal distance covered by the ball is$\mathbf{\text{130m}}$.

Write the formula to calculate the vertical distance covered by the baseball

role="math" localid="1663621792456" $\mathit{y}\mathbf{=}{\mathit{v}}_{\mathbf{y}}\mathit{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{g}{\mathit{t}}^{\mathbf{2}}$ ....... (I)

Here, $\mathit{y}$is the vertical distance covered by the baseball, ${\mathit{v}}_{\mathbf{y}}$is the component of the velocity in y direction and $\mathit{t}$is the time interval, $\mathit{g}$is the acceleration due to gravity.

calculate the time required to vertical distance covered by the baseball.

Write the formula to calculate the time required to vertical distance covered by the baseball

$t=\frac{d}{v_{\mathrm{x}}}$

Here,$v_{\mathrm{x}}$is the component of velocity in$x$direction and$d$is the horizontal distance covered by the baseball.

Substitute$\frac{d}{v_x}$for$t$in equation (I).

$y=v_{\mathrm{y}}\left(\frac{d}{v_{\mathrm{x}}}\right)-\frac{1}{2}g{\left(\frac{d}{v_{\mathrm{x}}}\right)}^2\left(\mathrm{III}\right)$

Write the expression for vertical component of the velocity

role="math" localid="1663622135911" $v_{\mathrm{y}}=v\sin \theta$

Here, $\theta$is the angle with the horizontal and $v$is the magnitude of velocity of the ball.

Write the expression for horizontal component of the velocity

$v_{\mathrm{x}}=v\cos \theta$

Substitute $v\cos \theta$for $v_{\mathrm{x}}$and $v\sin \theta$for $v_{\mathrm{y}}$in equation (II).

$\begin{array}{c}y=v\sin \theta \left(\frac{d}{v\cos \theta }\right)-\frac{1}{2}g{\left(\frac{d}{v\cos \theta }\right)}^2\\ =d\tan \theta -\frac{1}{2}g{\left(\frac{d}{v\cos \theta }\right)}^2\end{array}$

Substitute$41.7\text{m/s}$for$v,35.0^{\circ }$for$\theta ,130\text{m}$, for$d$and$9.81{\text{m/s}}^2$for$g$to find$y$.

$\begin{array}{c}y=130\mathrm{mtan}35.0°-\frac{1}{2}\times 9.81\mathrm{m}/{\mathrm{s}}^2\times {\left(\frac{130\mathrm{m}}{41.7\mathrm{m}/\mathrm{scos}35.0°}\right)}^2\\ =19.98\mathrm{m}\\ \approx 20.0\mathrm{m}\end{array}$

Therefore, the given situation is impossible because the calculated vertical distance is $20.0\mathrm{m}$lesser than the height of the bleaches.