A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms\(^{-2}\)]. Calculate the speed of project...

To determine the speed of projection of a body projected vertically upwards in a vacuum, we can use the principles of kinematics and the given information. Here are the steps:

1. Understand the total time of flight: The total time taken for the body to go up and come back down is 1.2 seconds.

2. Determine the time to reach the maximum height: Since the time to go up is equal to the time to come down, the time to reach the maximum height is half of the total time of flight.

   $t_{\text{up}}=\frac{1.2\text{ s}}{2}=0.6\text{ s}t\_\{\backslash text\{up\}\}\; =\; \backslash frac\{1.2\; \backslash text\{\; s\}\}\{2\}\; =\; 0.6\; \backslash text\{\; s\}$

3. Use the kinematic equation for velocity: The velocity at the maximum height is 0 (since the body momentarily stops before falling back down). We can use the equation:

   $v=u-gtv\; =\; u\; -\; gt$

   where $vv$ is the final velocity (0 m/s at the maximum height), $uu$ is the initial velocity (speed of projection), $gg$ is the acceleration due to gravity (10 m/s²), and $tt$ is the time taken to reach the maximum height (0.6 s).

   Setting $v=0v\; =\; 0$:

   $0=u-g\cdot t0\; =\; u\; -\; g\; \backslash cdot\; t$

4. Solve for $uu$:

   $u=g\cdot t=10{\text{ m/s}}^2\cdot 0.6\text{ s}=6\text{ m/s}u\; =\; g\; \backslash cdot\; t\; =\; 10\; \backslash text\{\; m/s\}^2\; \backslash cdot\; 0.6\; \backslash text\{\; s\}\; =\; 6\; \backslash text\{\; m/s\}$

Therefore, the speed of projection is $\boxed{{\displaystyle 6.0\text{ m/s}}}\backslash boxed\{6.0\; \backslash text\{\; m/s\}\}$6.0 m/s