You throw a glob of putty straight up toward the ceiling, which is 3.60 m...

Step 1
For freely falling bodies, the equations for the analysis model particle with constant acceleration also hold true.
The acceleration caused by gravity is the steady acceleration of a body falling freely.${\displaystyle (a_y=g=9.8\frac{m}{s^2})}$.
Since, we took the +ve direction to be upward, the acceleration is -ve downward ${\displaystyle (-a_y)}$
a) To calculate ${\displaystyle v_{yf}}$, we will use the following equation:
${\displaystyle v_{yf}^2=v_{yi}^2+2a_y(y_f-y_i)}$
${\displaystyle y_f-y_i=3.60m}$
${\displaystyle v_{yf}^2={9.50}^2+2(-9.8)\left(3.60\right)}$
${\displaystyle v_{yf}=4.44\frac{m}{s}}$
Step 2
b) We will use the following equation to determine the time:
$v_{yf}=v_{yi}+a_{y}t$
$4.44=9.50-9.8t$
${\displaystyle t=0.52s}$

Step 1
Given:
Distance, s=3.60 m
NSK
the initial speed of the putty, u = 9.50 m/s
Acceleration,
Step 2
a) We know that Gravitation acceleration acting downwards, so the value of the acceleration is a negative value
that is $f=-9.80m/s^2$
Let us take v be the final velocity,
$v^2=u^2+2fs$
Substitute the given value:

$\begin{array}{}v=9.50m/s,s=3.60m,f=-9.80m/s^2\\ v^2=(9.50{)}^2-2\times 9.80\times 3.60\\ v^2=90.25-70.56\\ v=\sqrt{19.69}\end{array}$
therefore, v=4.44m/s
Hence the speed of the putty just before it strikes the ceiling is 4.44m/s