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...

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 body at the maximum height of its motion

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 $\boxed{{\displaystyle 0.36\text{ J}}}\backslash boxed\{0.36\; \backslash text\{\; J\}\}$