A track meet is held on a planet in a distant solar system. A shot-putter releases a shot at a point 2.0mabove ground level. A stroboscopic plot of the position of the shot is shown in Fig. 4-61, where the readings areapart and the shot is released at time t = 0. (a) What is the initial velocity of the shot in unit-vector notation? (b) What is the magnitude of the free-fall acceleration on the planet? (c) How long after it is released does the shot reach the ground? (d) If an identical throw of the shot is made on the surface of Earth, how long after it is released does it reach the ground?

• Height at which the shot is released, $d_0=2.0\mathrm{m}.$
• Time interval between the readings,$t=0.5s.$

The kinematic equations are the set of equations of motion that gives the relationship between the initial velocity, final velocity, displacement, acceleration, and time of motion of the given object

Using kinematic equations, we can find the components of initial velocity and using those components we can write the initial velocity in vector form.

Using the same equation, we can find the acceleration of the planet and the time taken by a shot to reach the ground of the planet and earth.

Formula:

The second kinematic equation of motion,$d=d_0+v_{i}t+\left(\frac{1}{2}\right)a{t}^2$ (i)

Here,$v_i$ is the initial velocity, a is acceleration, t is time, and d is displacement.

From the graph, we can find that at t = 0 the position of the shot $\left(x_{i,}{y}_i\right)=(0,0)$is and at the total time $t_f=2.5s$is and the final position is $(x_f,y_f)=(25,2)$.

Using equation (i), we can find the x-component of the initial velocity. Substitute the given values in equation (i).

$$\begin{gathered} v_{ix}=\frac{x_f-x_i}{t_f-t_i}(\because a=0m/s^2) \\ =\frac{25m-0}{2.5s-0} \\ =10m/s \end{gathered}$$

Here, we take g’ as acceleration due to the gravity of the planet. Substitute the given values in equation (i).

role="math" localid="1657003784848" $$\begin{gathered} 2=2+v_{iy}t+\frac{1}{2}(-g\text{'})t^2 \\ {v}_{iy}t=\frac{1}{2}(-g\text{'})t^2 \\ {v}_{iy}=\frac{1}{2}(-g\text{'})t^1 \\ =\frac{1}{2}(2.5m/s^2)g\text{'} \\ =1.25g\text{'} \\ g\text{'}=\frac{v_{iy}}{1.25}..........................................\left(a\right) \end{gathered}$$

Similarly, we can find y-component of initial velocity as follows:

Using values distance.so the time is,$y_f=6.0mand{y}_i=2.0m$.so the time is, $t_f=2s$.Substitute the given values in equation (i).

$$\begin{gathered} 6.0m=2.0m+v_{iy}(2.0s)-\frac{1}{2}g\text{'}{(2.0s)}^2(6.0-2.0)=2v_{iy}+2g\text{'} \\ \frac{4}{2}=v_{iy}-g\text{'} \\ {v}_{iy}=2+g\text{'}...............................................\left(b\right) \\ {v}_{iy}=2+\left(\frac{v_{iy}}{1.25}\right)(\because g\text{'}=\frac{v_{iy}}{1.25}) \\ {v}_{iy}=\frac{2.5+v_{iy}}{1.25} \\ 1.25v_{iy-}{v}_{iy}=2.5 \\ 0.25v_{iy}=2.5 \\ {v}_{iy}=\frac{2.5}{0.25} \\ =10m/s \end{gathered}$$

Now the initial velocity in vector form can be written as:

$v_i=10\hat{i}+10\hat{j}$

Hence, the initial velocity is$10\hat{i}+10\hat{j}$ .

Comparing equations (a) and (b), we get the magnitude of free-fall acceleration as given:

$$\begin{gathered} 1.25g\text{'}=2+g\text{'} \\ 0.25g\text{'}=2 \\ g\text{'}=\frac{2}{0.25} \\ =8m/s^2 \end{gathered}$$

Hence, the value of free-fall acceleration is$8m/s^2$.

Using equation (i) and the value of free-fall acceleration, we can get the time taken by shot to reach ground. Substitute the values in equation (i).

$$\begin{gathered} 0=2+10t\text{'}-\frac{1}{2}\left(8\right)t^2 \\ t\text{'}=2.7s \end{gathered}$$

Hence, the value of the time taken by shot to reach the ground is 2.7s .

On ground the $g=9.8m/s^{2,}$,

Using equation (i) and the given values, we get the time taken by shot to reach earth. Substitute the given values in equation (i).

$$\begin{gathered} 0=2m+10m/s.t-(0.5)(9.8m/s^2)t^2 \\ t=2.2s \end{gathered}$$

Hence, the time taken by shot to reach the ground on earth is$2.2s$.