Question

part 1 of 3
a home run in a baseball game is hit in such a way that the ball just clears the wall which surrounds the outfield.
the acceleration of gravity is 9.8 m/s².
how long does it take the ball to reach the wall? neglect air friction.
answer in units of s.
part 2 of 3
what is the initial speed of the ball?
answer in units of m/s.

Explanation:

Step1: Analyze horizontal motion

The horizontal - motion equation is $x = v_{0x}t$, where $x$ is the horizontal displacement, $v_{0x}$ is the initial horizontal velocity, and $t$ is the time. The initial horizontal velocity $v_{0x}=v_0\cos\theta$, and $x = 121\ m$, $\theta = 44^{\circ}$. The vertical - motion equation for the vertical displacement $y - y_0$ is $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, where $y - y_0=(10.6 - 1)\ m = 9.6\ m$, $v_{0y}=v_0\sin\theta$, and $g = 9.8\ m/s^{2}$. From the horizontal motion $x = v_0\cos\theta\times t$, we have $v_0=\frac{x}{t\cos\theta}$.

Step2: Substitute $v_0$ into vertical - motion equation

Substitute $v_0=\frac{x}{t\cos\theta}$ into the vertical - motion equation $y - y_0 = v_0\sin\theta\times t-\frac{1}{2}gt^{2}$. We get $y - y_0=\frac{x}{t\cos\theta}\times\sin\theta\times t-\frac{1}{2}gt^{2}$. Since $\frac{\sin\theta}{\cos\theta}=\tan\theta$, the equation becomes $y - y_0=x\tan\theta-\frac{1}{2}gt^{2}$.

Step3: Solve for $t$ in part 1

Given $x = 121\ m$, $y - y_0=9.6\ m$, $\theta = 44^{\circ}$, and $g = 9.8\ m/s^{2}$. Substitute these values into $y - y_0=x\tan\theta-\frac{1}{2}gt^{2}$:

$$
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Step4: Solve for $v_0$ in part 2

Since $x = v_0\cos\theta\times t$, then $v_0=\frac{x}{t\cos\theta}$. Substitute $x = 121\ m$, $t = 4.68\ s$, and $\theta = 44^{\circ}$ ($\cos\theta=\cos(44^{\circ})\approx0.7193$):

$$ v_0=\frac{121}{4.68\times0.7193}=\frac{121}{3.3663}\approx35.95\ m/s $$

Answer:

part 1: $4.68\ s$
part 2: $35.95\ m/s$