91Ó°ÊÓ

The launching velocity of the ice ball relative to sled:$\left(V_{0s}\right)=V_{0x}\hat{i}+V_{0y}\hat{j}$

When two frames of reference Pand Qare traveling relative to each other at a constant velocity, the velocity of a particleAas measured by an observer in frame Pusually differs from that measured in frame Q. The relation between two measured velocities is,

${\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{A}\mathbf{P}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{A}\mathbf{Q}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{Q}\mathbf{P}}$

Here${\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{QP}}$ is the velocity of Q with respect to P.

Using the relative motion concept, we can find the magnitude and the direction of the resultant vector.

Formulae:

The second equation of the kinematic motion, $∆y=V_{y}t-\left(\frac{1}{2}\right)g{t}^2$ (i)

The displacement of a body, x=vt (ii)

The velocity of sled relative to the ground$\left(V_{sg}\right)$can be written in the vector notation as,

${\stackrel{â‡\P Ä}{V}}_{sg}=-V_s\hat{i}$

The launching velocity of the ice ball relative to sled,$\left(\stackrel{â‡\P Ä}{V_{os}}\right)=V_{0x}\hat{i}+V_{0y}\hat{j}$

So, the launching velocity of the ice ball relative to sled using the given equation can be written as follows:

$$\begin{gathered} \left(\stackrel{â‡\P Ä}{V_{os}}\right)=\left(V_{0x}-V_s\right)\hat{i}+V_{0y}\hat{j} \\ {V}_{ogy}=\left(V_{0x}-V_s\right)\mathrm{and} \\ {\mathrm{V}}_{\mathrm{ogy}}={\mathrm{V}}_{\mathrm{oy}} \end{gathered}$$

The horizontal displacement of the ball is given as:

$∆X_{bg}=\left(V_{0X}-V_s\right)t.....................\left(a\right)$

The vertical displacement of the ball is given suing equation (i) and the given data as:

$$\begin{gathered} ∆Y_{bg}=V_{oy}t-\left(\frac{1}{2}\right)g{t}^2 \\ 0=V_{oy}t-\left(\frac{1}{2}\right)g{t}^2 \\ {V}_{oy}t-\left(\frac{1}{2}\right)g{t}^2 \\ t=\frac{2V_{o{y}_{}}}{g}..............................\left(1\right) \end{gathered}$$

Putting this value for t in the equation (a) for$X_{og}$we get,

$∆X_{bg}=\frac{2V_{ox}{V}_{oy}}{g}-\frac{2V_{oy}{V}_s}{g}$

From graph, we can get the above value as follows:

$∆X_{bg}=40-4V_s...........................\left(b\right)$

Comparing these two equations (a) and (b) from part (a) calculations for, we can write that,

$$\begin{gathered} \frac{2V_{oy}}{g}=4 \\ {V}_{oy}=\frac{4g}{2} \end{gathered}$$

Substitute the given values.

$$\begin{gathered} V_{oy}=\frac{4\left(9.8\right)}{2} \\ =19.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, the velocity vector of the x-component can be calculated as:

$$\begin{gathered} \frac{2V_{0x}{V}_{oy}}{g}=40 \\ {V}_{0x}=\frac{40g}{2V_{0y}} \end{gathered}$$

Substitute the given values.

$$\begin{gathered} V_{0x}=\frac{40\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{2\left(19.6\mathrm{m}/\mathrm{s}\right)} \\ =10\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the velocity is 10 m/s.

So, from part (a) calculations, we can get that the value of the velocity is 19.6 m/s.

The distance$∆X_{bs}$is independent of the speed of the sled.

Thus, the distance value can be given using equations (ii) and (1) as follows:

$∆X_{bs}=10×\frac{2V_{oy}}{g}$

Substitute the given values.

$$\begin{gathered} ∆X_{bs}=10×\left(\frac{2×19.6\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^2}\right) \\ =10×4 \\ =40\mathrm{m} \end{gathered}$$

Hence, the value of the distance at this given speed is $40\mathrm{m}$.

The distance$∆X_{bs}$ is independent of the speed of the sled.

So, the value of distance at this given speed is 40 m.