Question

Problem 1. A batter hits a baseball so that it leaves the bat at speed vo = 37.0m/s at an angle αo = 53.1◦, at a location where g = 9.80m/s2.

Answer 1

(a) Let's first find $x$- and $y$-components of initial velocity of the baseball:

$$v_{0x}=v_0cos\alpha_0=37\ \dfrac{m}{s}\times cos53.1^{\circ}=22.2\ \dfrac{m}{s},$$ $$v_{0y}=v_0sin\alpha_0=37\ \dfrac{m}{s}\times sin53.1^{\circ}=29.6\ \dfrac{m}{s}.$$

Then, we can find the position of the baseball at $t=2\ s$:

$$x=x_0+v_{0x}t+\dfrac{1}{2}at^2=0+22.2\ \dfrac{m}{s}\times2\ s+0=44.4\ m,$$ $$y=y_0+v_{0y}t+\dfrac{1}{2}gt^2=0+29.6\ \dfrac{m}{s}\times2\ s+\dfrac{1}{2}\times()$$ $$y=0+29.6\ \dfrac{m}{s}\times2\ s+\dfrac{1}{2}\times(-9.8\ \dfrac{m}{s^2})\times(2\ s)^2=39.6\ m.$$

Therefore, the position of the baseball at $t=2\ s$ is (44.4 m, 39.6 m).

Let's find the magnitude and direction of the velocity of the baseball at $t=2\ s$.

The horizontal component of the velocity of baseball doesn't change during the flight:

$$v_{xf}=22.2\ \dfrac{m}{s}.$$

Let's find the vertical component of baseball's velocity:

$$v_{yf}=v_{0y}+gt=29.6\ \dfrac{m}{s}+(-9.8\ \dfrac{m}{s^2})\times2\ s=10\ \dfrac{m}{s}.$$

Finally, we can find the magnitude of the baseball's velocity at $t=2\ s$:

$$v_f=\sqrt{v_{xf}^2+v_{yf}^2}=\sqrt{(22.2\ \dfrac{m}{s})^2+(10\ \dfrac{m}{s})^2}=24.35\ \dfrac{m}{s}.$$

We can find the direction of the baseball's velocity from the geometry:

$$\alpha=tan^{-1}(\dfrac{v_{yf}}{v_{xf}})=tan^{-1}(\dfrac{10\ \dfrac{m}{s}}{22.2\ \dfrac{m}{s}})=24.3^{\circ}.$$

(b) We can find the time that the baseball takes to reach the highest point of its flight as follows:

$$v_{yf}=v_{0y}+gt,$$ $$0=v_{0y}+gt,$$ $$t=-\dfrac{v_{0y}}{g}=-\dfrac{29.6\ \dfrac{m}{s}}{-9.8\ \dfrac{m}{s^2}}=3.02\ s.$$

We can find the height of the baseball at this point as follows:

$$h=v_{0y}t+\dfrac{1}{2}gt^2,$$ $$h=29.6\ \dfrac{m}{s}\times3.02\ s+\dfrac{1}{2}\times(-9.8\ \dfrac{m}{s^2})\times(3.02\ s)^2=44.7\ m.$$

(c) We can find the horizontal range of the baseball as follows:

$$R=v_{0x}t_{flight}=v_{0x}2t=22.2\ \dfrac{m}{s}\times2\times3.02\ s=134\ m.$$