Question

1. a home run is hit in such a way that the baseball just clears a wall 21.0 m high, located 130 m from home plate. the ball is hit at an angle of 35.0° to the horizontal, and air resistance is negligible. find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (assume the ball is hit at a height of 1.00 m above the ground.)

Explanation:

Step1: Analyze horizontal motion

The horizontal motion is a uniform - motion. The horizontal displacement $x = v_{0x}t$, where $v_{0x}=v_0\cos\theta$ and $\theta = 35.0^{\circ}$, $x = 130\ m$. So $x = v_0\cos\theta\times t$, and $t=\frac{x}{v_0\cos\theta}$.

Step2: Analyze vertical motion

The vertical displacement $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, where $y - y_0=21.0 - 1.00=20.0\ m$, $v_{0y}=v_0\sin\theta$, and $g = 9.8\ m/s^{2}$. Substitute $t=\frac{x}{v_0\cos\theta}$ into the vertical - displacement equation: $y - y_0=v_0\sin\theta\times\frac{x}{v_0\cos\theta}-\frac{1}{2}g(\frac{x}{v_0\cos\theta})^{2}$.
Simplify the equation: $y - y_0=x\tan\theta-\frac{gx^{2}}{2v_0^{2}\cos^{2}\theta}$.
Rearrange the equation to solve for $v_0$:
\[

$$\begin{align*} \frac{gx^{2}}{2v_0^{2}\cos^{2}\theta}&=x\tan\theta-(y - y_0)\\ v_0^{2}&=\frac{gx^{2}}{2\cos^{2}\theta(x\tan\theta-(y - y_0))} \end{align*}$$

\]
Substitute $x = 130\ m$, $\theta = 35.0^{\circ}$, $y - y_0 = 20.0\ m$, and $g = 9.8\ m/s^{2}$:
\[

$$\begin{align*} v_0^{2}&=\frac{9.8\times130^{2}}{2\times\cos^{2}35.0^{\circ}(130\times\tan35.0^{\circ}-20.0)}\\ v_0^{2}&=\frac{9.8\times16900}{2\times0.6729(130\times0.7002 - 20.0)}\\ v_0^{2}&=\frac{165620}{2\times0.6729(91.026 - 20.0)}\\ v_0^{2}&=\frac{165620}{2\times0.6729\times71.026}\\ v_0^{2}&=\frac{165620}{95.37}\\ v_0&\approx41.6\ m/s \end{align*}$$

\]

Step3: Find the time $t$

Since $t=\frac{x}{v_0\cos\theta}$, substitute $x = 130\ m$, $v_0\approx41.6\ m/s$, and $\theta = 35.0^{\circ}$:
\[

$$\begin{align*} t&=\frac{130}{41.6\times\cos35.0^{\circ}}\\ t&=\frac{130}{41.6\times0.8192}\\ t&\approx3.80\ s \end{align*}$$

\]

Step4: Find the velocity components at the wall

The horizontal component of velocity remains constant: $v_x = v_0\cos\theta=41.6\times0.8192\approx34.1\ m/s$.
The vertical component of velocity is $v_y=v_0\sin\theta - gt$. Substitute $v_0 = 41.6\ m/s$, $\theta = 35.0^{\circ}$, $g = 9.8\ m/s^{2}$, and $t = 3.80\ s$:
\[

$$\begin{align*} v_y&=41.6\times0.5736-9.8\times3.80\\ v_y&=23.86 - 37.24\\ v_y&\approx - 13.4\ m/s \end{align*}$$

\]
The speed of the ball at the wall is $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{34.1^{2}+(- 13.4)^{2}}=\sqrt{1162.81 + 179.56}=\sqrt{1342.37}\approx36.6\ m/s$.

Answer:

(a) $v_0\approx41.6\ m/s$
(b) $t\approx3.80\ s$
(c) $v_x\approx34.1\ m/s$, $v_y\approx - 13.4\ m/s$, $v\approx36.6\ m/s$