To find the time the baseball was in the air, we can use the equation of motion:
$$h = v_0 \sin(\theta) t - \frac{1}{2} g t^2$$
where:
- $h$ is the maximum height reached by the baseball,
- $v_0$ is the initial velocity of the baseball,
- $\theta$ is the angle of the throw,
- $t$ is the time spent in the air, and
- $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$).
Given that $v_0 = 23 \, \text{m/s}$ and $\theta = 25^\circ$, we can solve for $t$.
We can rearrange the equation to solve for $t$:
$$t = \frac{v_0 \sin(\theta) \pm \sqrt{v_0^2 \sin^2(\theta) + 2gh}}{g}$$
In this case, since we are looking for the time when the ball was in the air (i.e., at the maximum height), $h = 0$.
Therefore, the equation simplifies to:
$$t = \frac{2v_0 \sin(\theta)}{g}$$
Substituting the given values, we have:
$$t = \frac{2(23 \, \text{m/s})\sin(25^\circ)}{9.8 \, \text{m/s}^2}$$
Simplifying further, we get:
$$t \approx 2.92 \, \text{s}$$
So the ball was in the air for approximately 2.92 seconds.
To find the maximum height reached by the ball, we can use the equation:
$$h = v_0 \sin(\theta) t - \frac{1}{2} g t^2$$
Since we already know that the time $t$ is approximately 2.92 seconds and $h = 0$, the equation simplifies to:
$$0 = v_0 \sin(\theta) (2.92) - \frac{1}{2} g (2.92)^2$$
Solving for $v_0 \sin(\theta)$:
$$v_0 \sin(\theta) = \frac{1}{2} g (2.92)^2$$
Substituting the known values:
$$v_0 \sin(\theta) \approx \frac{1}{2} (9.8 \, \text{m/s}^2)(2.92 \, \text{s})^2$$
Simplifying:
$$v_0 \sin(\theta) \approx 42.77 \, \text{m}$$
Therefore, the ball traveled approximately 42.77 meters above the thrower.
A baseball is thrown at an angle of 25° relative to the ground at a speed of 23 m/s. How long was it in the air? How high above the thrower did the ball travel?
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