The pitcher in a slow-pitch softball game releases the ball at a pointabove ground level. A stroboscopic plot of the position of the ball is shown in Fig. 4-60, where the readings areapart and the ball is released at. (a) What is the initial speed of the ball? (b) What is the speed of the ball at the instant it reaches its maximum height above ground level? (c) What is that maximum height?

• Initial height at which the ball is above ground level,$d_0=3\mathrm{ft}$.
• Time interval between the readings, $t=0.25\mathrm{s}$.

Kinematic equations are the equations of motion that relate initial velocity, final velocity, acceleration, displacement, and time.

We can find the initial speed of the ball using the kinematic equation of distance. Using time and respective distance we can find the speed at maximum height and maximum height.

Formulae:

The second kinematic equation of motion,$d^d=d_o+V_{i}t+\left(\frac{1}{2}\right)a{t}^2$ (i)

Here,$d$is final displacement,$d_0$is initial displacement,$V_i$is initial velocity,$t$is time and$a$is acceleration.

From the graph, we can find the total time is $t=1.25\mathrm{s}$and the horizontal distance is$40\mathrm{ft}$.

We can find the horizontal component of initial velocity, $a_x=0$along the horizontal direction using equation (i). Substitute the given values in equation (i).

role="math" localid="1657023194827" $\begin{array}{l}40\mathrm{ft}=0+{\mathrm{V}}_{\mathrm{ix}}(1.25\mathrm{s})\\ V_{ix}=\frac{40\mathrm{ft}}{1.25\mathrm{s}}\\ =32\mathrm{ft}/\mathrm{s}\end{array}$

Similarly, we can find the initial vertical velocity component, ball come at the same level by traveling vertical distance using equation (i). Therefore, substitute the given values in equation (i) to calculate the initial vertical velocity.

role="math" localid="1657023168439" $$\begin{gathered} 3=3+{\mathrm{V}}_{\mathrm{iy}}(1.25\mathrm{s})+(0.5)(-32\mathrm{ft}/{\mathrm{s}}^2){(1.25s)}^2 \\ 0={\mathrm{V}}_{\mathrm{iy}}(1.25\mathrm{s})-25\mathrm{ft} \\ {\mathrm{V}}_{\mathrm{iy}}=\frac{25\mathrm{ft}}{1.25\mathrm{s}} \\ =20\mathrm{ft}/\mathrm{s} \end{gathered}$$

Hence, the magnitude of the initial speed is calculated by substituting the value of vertical and horizontal velocity in the equation for the magnitude.

$\begin{array}{l}{V}_i=\sqrt{{(32\mathrm{ft}/\mathrm{s})}^2++{(20\mathrm{ft}/\mathrm{s})}^2}\\ =37.73\mathrm{ft}/\mathrm{s}\\ \approx 38\mathrm{ft}/\mathrm{s}\end{array}$

Hence, the value of initial speed is $38\mathrm{ft}/\mathrm{s}$.

At maximum height vertical velocity becomes $V_y=0$and there is only horizontal velocity component so the speed at the maximum height is $V_x=32\mathrm{ft}/\mathrm{s}$.

Here we can use the vertical velocity component and half time of flight to find maximum height. Substituting the values in equation (i), we get

$$\begin{gathered} y=3\mathrm{ft}+(20\mathrm{ft}/\mathrm{s})(0.625\mathrm{s})+(0.5)(-32\mathrm{ft}/{\mathrm{s}}^2){(0.625\mathrm{s})}^2 \\ =3\mathrm{ft}+12.5\mathrm{ft}-6.25\mathrm{ft} \\ =9.25\mathrm{ft} \\ \approx 9.3\mathrm{ft} \end{gathered}$$

Hence, the value of the maximum height of the ball is 9.3 ft.