Question

Figure 2-35 shows the speed v versus height y of a ball tossed directly upward, along a y axis. Distance d is0.40 m. The speed at height${\mathit{y}}_{\mathbf{A}}$is${\mathit{v}}_{\mathbf{A}}$.The speed at height${\mathit{y}}_{\mathbf{B}}$is$\frac{\mathbf{1}}{\mathbf{3}}{\mathit{V}}_{\mathbf{A}}$. What is speed${\mathit{V}}_{\mathbf{A}}$?

Short answer

The speed of a ball at height $y_A$is ${\mathrm{V}}_{\mathrm{A}}=3.0\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given Data

Given,

$V_B=\frac{1}{3}{V}_A$

$y_B-y_A=0.40m$

Understanding the concept of motion with constant acceleration

This problem is based on kinematics in which the motion of an object can be

determined with constant acceleration. The speed at a given location can be

found by using a kinematic equation.

Formula

$$\begin{gathered} v_f^2=v_i^2+2as \\ \left(\mathrm{i}\right) \end{gathered}$$

Calculation of speed

Rewrite the equation (i) using the given data.

$v_B^2=V_A^2+2\left(g\right)\left(y_B-y_A\right)$ (ii)

For the vertical motion,$g=-9.8m/s^2$

Substitute the values in equation (ii).

$$\begin{gathered} v_B^2=V_A^2+2\left(-98\right)\left(0.40\right) \\ \frac{1}{9}{V}_A^2=V_A^2-7.84 \\ 7.84=V_A^2-\frac{V_A^2}{9} \\ 7.84=\frac{8}{9}{V}_A^2 \\ 8.82=V_A^2 \\ {V}_A=2.969m/s \\ \approx 3m/s \end{gathered}$$

Therefore, the value of $v_A$is 3 m/s .