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?
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.