Answer
(a) The slope of the straight line is $0.0510\frac{s^2}{m}$.
(b) For a given release height the final speed of the tennis ball with air resistance is lower than the final speed of the tennis ball without air resistance.
(c) The graph of $ y-y_{0}$ vs $v_{y}^2$ will not be a straight line.
Work Step by Step
Let the origin of our system be at the various points of release and we take the (+)ve y-direction to be downwards.
We ignore the air resistance for the steel ball because the effect of air resistance experienced by the steel ball is less than that of the tennis ball.
The motion of the steel ball can be assumed with constant acceleration, we can apply the constant acceleration kinematics equation.
(a) Then,
$2a_{y}(y-y_{0})=v_{y}^2-v_{0y}^2$
$=> y-y_{0}=\frac{v_{y}^2}{2g}$->(1) , [since, $v_{0y}=0$ and $a_{y}=g$]
Therefore, the graph of $ y-y_{0}$ vs $v_{y}^2$ is a straight line with slope $\frac{1}{2g}=0.0510\frac{s^2}{m}$.
Thus, the slope of the straight line is $0.0510\frac{s^2}{m}$.
(b) For the tennis ball, the effect of the air resistance cannot be ignored. The presence of air resistance decreases the magnitude of the downward acceleration and the effect of air resistance increases as the speed of the object increases.
So, the motion of the tennis ball cannot be assumed with constant acceleration.
Thus, for a given release height the final speed of the tennis ball with air resistance is lower than the final speed of the tennis ball without air resistance.
(c) With air resistance the downward acceleration is less than $9.80\frac{m}{s^2}$ and the downward acceleration will vary with the speed of the ball.
So, the graph of $ y-y_{0}$ vs $v_{y}^2$ will not be a straight line.
