A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms-2]. Determine the potential energy of the...

Answer Details

To determine the potential energy of the body at the maximum height of its motion, we need to follow these steps:

1. Calculate the speed of projection (initial velocity): From the previous solution, we determined that the speed of projection $uu$ is 6 m/s.

2. Determine the maximum height: Use the kinematic equation for the vertical motion to find the maximum height:

   $v^2=u^2-2ghv^2\; =\; u^2\; -\; 2gh$

   At the maximum height, the final velocity $vv$ is 0, so:

   $0=u^2-2gh⟹h=\frac{u^2}{2g}0\; =\; u^2\; -\; 2gh\; \backslash implies\; h\; =\; \backslash frac\{u^2\}\{2g\}$

   Substituting $u=6\text{ m/s}u\; =\; 6\; \backslash text\{\; m/s\}$ and $g=10{\text{ m/s}}^{2}g\; =\; 10\; \backslash text\{\; m/s\}^2$:

   $h=\frac{(6\text{ m/s}{)}^2}{2\times 10{\text{ m/s}}^2}=\frac{36{\text{ m}}^2\mathrm{/}{\text{s}}^2}{20{\text{ m/s}}^2}=1.8\text{ m}h\; =\; \backslash frac\{(6\; \backslash text\{\; m/s\})^2\}\{2\; \backslash times\; 10\; \backslash text\{\; m/s\}^2\}\; =\; \backslash frac\{36\; \backslash text\{\; m\}^2/\backslash text\{s\}^2\}\{20\; \backslash text\{\; m/s\}^2\}\; =\; 1.8\; \backslash text\{\; m\}$

3. Calculate the potential energy (PE): The potential energy at the maximum height is given by:

   $\text{PE}=mgh\backslash text\{PE\}\; =\; mgh$

   Convert the mass from grams to kilograms:

   $m=20\text{ g}=0.02\text{ kg}m\; =\; 20\; \backslash text\{\; g\}\; =\; 0.02\; \backslash text\{\; kg\}$

   Now substitute $m=0.02\text{ kg}m\; =\; 0.02\; \backslash text\{\; kg\}$, $g=10{\text{ m/s}}^{2}g\; =\; 10\; \backslash text\{\; m/s\}^2$, and $h=1.8\text{ m}h\; =\; 1.8\; \backslash text\{\; m\}$:

   $\text{PE}=0.02\text{ kg}\times 10{\text{ m/s}}^2\times 1.8\text{ m}=0.36\text{ J}\backslash text\{PE\}\; =\; 0.02\; \backslash text\{\; kg\}\; \backslash times\; 10\; \backslash text\{\; m/s\}^2\; \backslash times\; 1.8\; \backslash text\{\; m\}\; =\; 0.36\; \backslash text\{\; J\}$

Therefore, the potential energy of the body at the maximum height of its motion is $\boxed{{\displaystyle 0.36\text{ J}}}\backslash boxed\{0.36\; \backslash text\{\; J\}\}$