Question

A fly ball is hit with an angle of 80 degrees with a initial speed of 40.0 m/s. How long does the fielder have to get underneath the ball? And what is the maximum height attained by the ball?

Answer 1

Question #36178

A fly ball is hit with an angle of 80 degrees with a initial speed of

40.0 m/s. How long does the fielder have to get underneath the ball? And

what is the maximum height attained by the ball?

Solution:

Let

$$v_0 = 40 \, \text{m/sec}$$ $$\alpha = 80{}^\circ$$ $$t = ?, H = ?$$

$H = v_{y0} t_h - \frac{1}{2} g t_h^2$ where $t_h$ is the time of moving upward, $v_{y0}$ is the vertical compound of initial velocity

$$v_y = v_{y0} - g t_h$$

Such as the velocity in the highest point is equal to zero

$$v_{y0} = g t_h, \quad t_h = \frac{v_{y0}}{g}$$ $$H = v_{y0} \frac{v_{y0}}{g} - \frac{1}{2} g \left(\frac{v_{y0}}{g}\right)^2$$ $$H = \frac{1}{2} \frac{v_{y0}^2}{g}$$

The full time is $t = 2t_h$

$$t = 2 \frac{v_{y0}}{g}$$ $$v_{y0} = v_0 \sin \alpha$$ $$H = \frac{1}{2} \frac{(v_0 \sin \alpha)^2}{g}$$ $$t = 2 \frac{v_0 \sin \alpha}{g}$$ $$t = 2 \frac{40 \cdot \sin 80}{9.8} = 8 \, \text{sec}$$ $$H = \frac{1}{2} \frac{(40 \cdot \sin 80)^2}{9.8} = 79.2 \, \text{m}$$

Answer: 8 sec, 79.2 m.