Problem :
A wooden block of mass $5 kg$. fell down from a tower on a swimming pool. The tower is of height $19.6 metre$ above the water surface in the pool. After it collides with the water's surface it moved inside the water vertically downwards with uniform acceleration of magnitude $4 m/sec^2$ , then it covered a distance $4.5 m$. before it starts returning back to float on the water. Find the magnitude of change of momentum of the wooden block due to its impact with the water surface in the pool.
So it should look like this :
My try :
1 - We get $V_1$ using $V^2 = V_o^2 + 2gd$ :
$d = -19.6$
, $V_o = 0$
, $g = -9.8m/s^2$
$V_1^2 = 0 + 2 \times -9.8 \times -19.6 = 384.16$
$V_1 = 19.6m/s$
2 - for $V_2$ i use $V^2 = V_o^2 + 2ad$ again :
$d = -4.5$
, $V_3 = 0$
, $a = -4m/s^2$
$0 = V_2^2 + 2 \times -4 \times -4.5 = V_2 + 36$
$V_2^2 = -36$
and then i am stuck here because i cannot take the root of a negative number , did i mess directions up ? i have no idea what went wrong , any help will be appreciated
thanks in advance.
Falling from height $h=19.6\,\text{m}$, the bock reaches the surface with $v=\sqrt{2ah}$ where $a=g=9.81\,\text{m}\text{s}^{-2}$. In the water, we have "inverted falling" as the overall acceleration is upward. Neverthelessm the same formula applies, but this time with $h=4.5\,\text{m}$ and $a=4\,\text{m}\text{s}^{-2}$. The difference between these to velocities at the water surface, multiplied by the mass, is the desired change in momentum.