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 equation is $x = v_{0x}t$, where $v_{0x}=v_0\cos\theta$ and $x = 130$ m, $\theta = 35.0^{\circ}$. So $x = v_0\cos\theta\times t$.

Step2: Analyze vertical motion

The vertical - motion equation is $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, where $y - y_0=21.0 - 1.00=20.0$ m, $v_{0y}=v_0\sin\theta$, and $g = 9.8$ m/s². So $y - y_0=v_0\sin\theta\times t-\frac{1}{2}gt^{2}$.
From $x = v_0\cos\theta\times t$, we can get $t=\frac{x}{v_0\cos\theta}$. Substitute $t$ into the vertical - motion equation:
\[

$$\begin{align*} y - y_0&=v_0\sin\theta\times\frac{x}{v_0\cos\theta}-\frac{1}{2}g(\frac{x}{v_0\cos\theta})^{2}\\ y - y_0&=x\tan\theta-\frac{gx^{2}}{2v_0^{2}\cos^{2}\theta} \end{align*}$$

\]
Substitute $x = 130$ m, $y - y_0 = 20.0$ m, $\theta = 35.0^{\circ}$, and $g = 9.8$ m/s² into the above equation:
\[

$$\begin{align*} 20.0&=130\times\tan35.0^{\circ}-\frac{9.8\times130^{2}}{2v_0^{2}\cos^{2}35.0^{\circ}}\\ 20.0&=130\times0.700-\frac{9.8\times16900}{2v_0^{2}\times0.672}\\ 20.0&=91-\frac{82810}{v_0^{2}\times0.672}\\ \frac{82810}{0.672v_0^{2}}&=91 - 20.0\\ \frac{82810}{0.672v_0^{2}}&=71\\ v_0^{2}&=\frac{82810}{0.672\times71}\\ v_0^{2}&=\frac{82810}{47.712}\\ v_0&\approx41.7\text{ m/s} \end{align*}$$

\]

Step3: Find the time to reach the wall

Since $t=\frac{x}{v_0\cos\theta}$, substitute $x = 130$ m, $v_0 = 41.7$ m/s, and $\theta = 35.0^{\circ}$:
\[
t=\frac{130}{41.7\times\cos35.0^{\circ}}=\frac{130}{41.7\times0.819}\approx3.8\text{ s}
\]

Step4: Find the velocity components at the wall

The horizontal - velocity component $v_x=v_0\cos\theta=41.7\times\cos35.0^{\circ}\approx41.7\times0.819 = 34.2$ m/s.
The vertical - velocity component $v_y=v_0\sin\theta - gt$, substitute $v_0 = 41.7$ m/s, $\theta = 35.0^{\circ}$, $g = 9.8$ m/s², and $t = 3.8$ s:
\[
v_y=41.7\times\sin35.0^{\circ}-9.8\times3.8=41.7\times0.574 - 37.24=24.0 - 37.24=- 13.2\text{ m/s}
\]
The speed of the ball at the wall $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{34.2^{2}+(-13.2)^{2}}=\sqrt{1169.64 + 174.24}=\sqrt{1343.88}\approx36.7$ m/s

Answer:

(a) $41.7$ m/s
(b) $3.8$ s
(c) $v_x = 34.2$ m/s, $v_y=-13.2$ m/s, $v\approx36.7$ m/s