Speed of a pitcher

Whether we let the ball fall freely or we throw it horizontally, if it start from the same height, the ball will hit the ground at the same time. Thus, we can use this formula to find out how much time the ball will take to hit the ground:

t = $\sqrt{2h/g}$

where t is time in seconds, h is height and g is gravity. In our case, the value is 0.6388765 s .

We throw the ball at a certain speed and the ball hits the ground 20 meters away from where we're standing. In other words, with its current speed the ball is able to reach 20 meters in 0.6388765 second. If speed (velocity) is v , distant reached by the ball is s and the time needed is t , we can express that:

s = v . t

v = $\frac{s}{t}$ = $\frac{20 m}{0.6388765 s}$

v = 31.3 m/s

First find the time it takes for the back to hit the ground. the only field affecting it in the vertical direction is gravity. We can use:

$y = v_it + {1 \over 2}at^2 \quad \Rightarrow \quad 2 = 0 + {1 \over 2}\cdot 9.8t^2 \quad \Rightarrow \quad 2 = 4.9t^2 \\ \quad t = \sqrt{2\over4.9} \quad \Rightarrow \quad t \approx 0.6388 \\ _{\text{Note: }v_i \text{ is zero because there is no initial velocity in the y component.}}$

Now we have the time it took the ball to reach travel 2m and hit the ground. This is the same time that it takes the ball to reach 20m after leaving our hands. We can use a similar formula:

$x = v_it +{1 \over 2}at \quad \Rightarrow \quad 20 = v_i\cdot 0.6388 + 0 \quad \Rightarrow \quad \\ v_i = {20 \over 0.6388} \quad \Rightarrow \quad v_i \approx 31.308 \\ _{\text{Note: There is no acceleration in the x component but it does have an initial velocity. }}$

Define two axes, $x$ and $y$ . The axes cross each other in the center of gravity of the ball. $x$ is vertical to the ground and $y$ is horizontal to the ground. $x$ decreases as the ball falls, while $y$ increases horizontally in the direction in which the ball is thrown. Now we must realize that we are looking for the speed $v$ along the $y$ -axis when the ball leaves your hand, which is at $t = 0s$ . This means we must find $v_{0_{\Large y}}$ .

$\large s_x(t) = v_{ \normalsize 0_{\Large x}}\large t + \frac{1}{2}\vec{g}t^2 \normalsize \textrm{ where } \large v_{ \normalsize 0_{\Large x}} = \normalsize 0 \space ms^{-1} \textrm{ and } \vec{g} = -9.8 ms^{-2}$

$\large s_y(t) = v_{ \normalsize 0_{\Large y}}\large t + \frac{1}{2}\vec{g}t^2 \normalsize \textrm{ where } \vec{g} = 0 \space ms^{-2}$
$\textrm{ and the value of } \large v_{ \normalsize 0_{\Large y}} \normalsize \textrm{ is to be found.}$

$s_x(t) = -2m \implies \frac{1}{2} \times -9.8t^2 = -2m \implies t = \sqrt{ \frac{-2 \times 2}{-9.8} } = 0.64 s$
$s_y(t) = 20m \implies v_{0_{\Large y}} \normalsize \times t = 20m \textrm{ where } t = 0.64s$

v_{0_{\Large y}} \normalsize = \frac{20}{0.64} = \fbox{ \$31 ms^{-1}\$ }

The ball follows the path of a horizontal projectile. Here, range (R) = 20 m height(H) = 2 m acceleration due to gravity(g) = 9.8 m/s^2 velocity(u) = ? R = u\sqrt{(2h)/g} On putting the values, u = 31.3 m/s

$y_{o}=2m$ $y=0$

$y=y_{o}+v_{oy}t-\frac{1}{2}gt^{2}$ $0=2+0-\frac{1}{2}9.8t^{2}$ $2=4.9t^{2}$ $t^{2}=\frac{2}{4.9}$ $t=\sqrt{\frac{2}{4.9}}$ $t=\frac{2}{7}\sqrt{5}$ $t=0,638 s$

Then in horizontal;

$x=v_{x}t$ $20=v \times 0,638$ $v=\frac{20}{0,638}$ $v=31,3 m/s$