Hi there!
A.
Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.
Thus, the time to its highest point:
Now, we can determine the velocity at which the can was launched at using the following equation:
In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.
Therefore:
***vsinθ is the vertical component of the velocity.
Solve for 'v':
Now, recall that:
Plug in the expression for velocity:
B.
We can use the same process as above, where T' = 2T and Th = T.
C.
The work done in part B is 4 times greater than the work done in part A.