Question

A home run is hit in such a way that the baseball just clears a wallhigh, locatedfrom home plate. The ball is hit at an angle of$\mathbf{35}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}$to the horizontal, and air resistance is negligible. (a) Find the initial speed of the ball (b) the time it takes the ball to reach the wall(c) the velocity components and the speed of the ball when it reaches the wall. (Assume the ball is hit at a height ofabove the ground.)

Short answer

(a) The initial speed of ball is$41.7\text{m/s}$ .

(b) The time taken by the ball to reach the wall is $3.81 \text{s}$.

(c) The horizontal component of velocity is $34.1\text{m/s}$, the vertical component of velocity is$-13.4\text{m/s}$ and the speed of the ball when it reaches the wall is$36.7\text{m/s}$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

The height of wall is $21.0 \text{m}$.

The distance of wall from home is$130 \text{m}$ .

The ball hit the wall at the height from the ground is$1.00 \text{m}$ .

The ball hit the wall at angle of $35.0^0$.

The formula to calculate x component of initial velocity

Write the formula to calculate x component of initial velocity

${\mathit{v}}_{\mathbf{0}\mathbf{x}}\mathbf{=}{\mathit{v}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$

Here, $v_0$is the initial velocity, $v_{0x}$is the initial horizontal velocity and $\theta$is the angle of projection.

Substitute $35.0^0$for $\theta$in above equation to find$v_{0x}$ .

$\begin{array}{rcl}{v}_{0x}& =& v_0\cos \left({35.0}^0\right)\\ & =& 0.819v_0\end{array}$

calculate y component of initial velocity

Write the formula to calculate ycomponent of initial velocity

${\mathit{v}}_{\mathbf{0}\mathbf{y}}\mathbf{=}{\mathit{v}}_{\mathbf{0}}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

Here, $v_{0y}$is the initial velocity.

Substitute $35.0^0$for $\theta$in above equation to find $v_{0y}$.

$v_{0\text{y}}=0.573v_0$

Write the formula to calculate time

$t=\frac{d}{v_{v_{\text{ax}}}}=\frac{d}{0.819v_0}$

Here, d is the distance of wall.

Calculate the height of the wall above the home plate.

$\begin{array}{rcl}h& =& 21 \text{m}-1 \text{m}\\ & =& 20 \text{m}\\ & & \end{array}$

Step 4(a): Determination of the initial speed of ball 

Write the kinematic equation for distance travelled

$h=v_{0y}t+\frac{1}{2}a{t}^2$

Here, a is the acceleration and t is the time.

Substitute$0.573v_0$ for$v_{0y},-9.8{\text{m/s}}^2$ for $a,20 \text{m}$for $h,\frac{d}{0.81v_0}$for t and 130 m for d in above equation to find v₀.

$\begin{array}{rcl}20 \text{m}& =& \left(0.573v_0\right)\left(\frac{d}{0.819v_0}\right)+\frac{1}{2}\left(-9.8{\text{m/s}}^2\right){\left(\frac{d}{0.819v_0}\right)}^2\\ & =& \left(0.573v_0\right)\left(\frac{130 \text{m}}{0.819v_0}\right)+\frac{1}{2}\left(-9.8{\text{m/s}}^2\right){\left(\frac{130 \text{m}}{0.819v_0}\right)}^2\\ & =& (90.95 \text{m})-\frac{123456.79{\text{m}}^3{\text{/s}}^2}{v_0^2}\\ \frac{123456.79{\text{m}}^3{\text{/s}}^2}{v_0^2}& =& 70.95 \text{m}\\ & & \end{array}$

Further, solve to find .

$\begin{array}{rcl}\frac{123456.79{\text{m}}^3{\text{/s}}^2}{v_0^2}& =& 70.95 \text{m}\\ & =& \sqrt{\frac{123456.79{\text{m}}^3{\text{/s}}^2}{70.95 \text{m}}}\\ & =& 41.7 \text{m/s}\end{array}$

Therefore, the initial speed of ball is 41.7 m/s .

Step 5(b): The time taken by the ball to reach the wall

(b)

Write the equation for time using (III)

$t=\frac{d}{0.819v_0}$

Substitute 41.7 m/s for v₀ and 130 m for d in equation (III) to find t.

$\begin{array}{rcl}t& =& \frac{130 \text{m}}{(0.819)(41.7 \text{m/s})}\\ & =& 3.81 \text{s}\\ & & \end{array}$

Therefore, the time taken by the ball to reach the wall is 3.81 s.

Step 6(c): calculate speed of the ball

Write the formula to calculate vertical component of velocity

$v_{0y}^{\text{'}}=v_{0y}-gt$

Here, localid="1663777203595" $v_{0y}^{\text{'}}$is the final vertical velocity.

Substitute $0.573v_0$for $v_{0y}$in the above equation

$v_{0\text{y}}^{\text{'}}=0.573v_0-gt$

Write the formula to calculate speed of the ball when it reaches the wall

$v=\sqrt{v_{0\text{x}}^2+v_{0\text{y}}^2}$

Substitute 41.7 m/s for v₀ in equation (I) to find $v_{0\text{x}}$.

$\begin{array}{rcl}{v}_{0\text{x}}& =& (0.819)(41.7 \text{m/s})\\ & =& 34.1 \text{m/s}\\ & & \end{array}$

Substitute 41.7 m/s for v₀ , 9.8 m/s²for g and 3.81 s for t in equation (IV) to find ${v^{\text{'}}}_{0y}$.

$\begin{array}{rcl}{v}_{0\text{y}}^{\text{'}}& =& (0.573)(41.7 \text{m/s})-\left(9.8 {\text{m/s}}^{\text{2}}\right)(3.81 \text{s})\\ & =& -13.4 \text{m/s}\\ & & \end{array}$

Substitute 34.1 m/s for$v_{0\text{x}}$ and -13.4 m/s for${v^{\text{'}}}_{0y}$ in equation .

$\begin{array}{rcl}v& =& \sqrt{{(34.1 \text{m/s})}^2+{(-13.4 \text{m/s})}^2}\\ & =& 36.7 \text{m/s}\\ & & \end{array}$

Therefore, the horizontal component of velocity is 34.1 m/s , the vertical component of velocity is -13.4 m/s and the speed of the ball when it reaches the wall is 36.7 m/s.