More constant acceleration questions

[kylera]

Because there's no solution, and my book rated this problem as one of the tougher ones compared to the others, I have no idea whether I did this right, but my gut feeling is that it's wrong because I solved it waaaay too quickly.

Question: Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upoward at a height of 1.5m above the ground. When you quickly move the nozzle away from the vertical, you hear the water striking the ground next to you for another 2.0s. what is the water speed as it leaves the nozzle?

My solution: Use the formula that determines position with time, initial velocity and acceleration:
x = x(initial) + v(initial)*time + .5*a*t^2
t = 2.0s, a = -9.8m/s^2
-1.5 = 0 + 2v - 2*9.8
v(initial) = 9.1 m/s

Any pointers?

 

[lightgrav]

okay, I got 9.05 m/s .
maybe the author thought it was "difficult" because he did not include one exactly like it (solved as an example).
Authors don't distinguish between "challenging", as opposed to "long" or "tedious".

 

[quicknote]

As a scientist, it is important to always question your solutions and look for potential errors or discrepancies. In this case, your solution seems reasonable and the steps you took to solve the problem are correct. However, it is always a good idea to check your answer using different methods or equations to confirm its accuracy. Additionally, you can also try plugging in different values for the initial velocity and acceleration to see if they still satisfy the equation. If they do not, then it is likely that your solution is incorrect. It is also helpful to double-check your calculations and make sure that all units are consistent throughout the problem. Keep in mind that practice and patience are key in solving tough problems, so don't be discouraged if it takes some time to find the correct solution.