Answer
See the detailed answer below.
Work Step by Step
a) First, we need to sketch the experiment, as we see below.
We know that once the ball leaves the edge of the table will be under the free-fall acceleration and no horizontal forces are exerted on it (air resistance here is neglected).
Thus, the range is given by
$$R=\Delta x=v_{ix}t+\frac{1}{2}\overbrace{a_x}^{=0}t^2$$
Thus,
$$R= v_{ix}t\tag 1$$
So we have to find the time of the trip of the ball from the edge of the table to the ground and we also need to find its releasing velocity in terms of $h$, $m$, and $J_x$.
The time could be found from the vertical kinematic formula of displacement.
$$\overbrace{y}^{=0}=\overbrace{y_i}^{=h}+\overbrace{v_{iy}}^{=0}t+\frac{1}{2}\overbrace{a_y}^{-g}t^2$$
$$0=h-\frac{1}{2}gt^2$$
Hence,
$$t=\sqrt{\dfrac{2h}{g}}\tag 2$$
Now we need to find the ball's releasing velocity and we know, from the impulse-momentum principle, that the impulse exerted on the ball equals the change in its momentum.
$$J_x=\Delta p_x=mv_{ix}-m\overbrace{v_{0x}}^{=0} =mv_{ix}$$
The ball was initially at rest before the spring impulse was exerted on it.
Thus, the releasing velocity is given by
$$v_{ix}=\dfrac{J_x}{m}\tag 3$$
Plug (2) and (3) into (1);
$$\boxed{R= \dfrac{J_x}{m} \left(\sqrt{\dfrac{2h}{g}}\right)}$$
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b) We have two variables here, in the boxed formula, which are the range and the mass of the ball. The height is constant, the spring impulse is constant, and the free-fall acceleration is constant.
To get the slope that gave us the impulse from the boxed formula above, we need to draw $\dfrac{1}{m}$ versus $R$
$$R=\overbrace{J_x\left(\sqrt{\dfrac{2h}{g}}\right)}^{slope} \dfrac{1}{m} $$
and yet the slope of the line will be given by
$${\rm Slope}=J_x\left(\sqrt{\dfrac{2h}{g}}\right)$$
Thus,
$$J_x=\dfrac{{\rm Slope}}{\sqrt{\dfrac{2h}{g}}}$$
$$\boxed{ J_x= {\rm Slope} \left(\sqrt{\dfrac{g}{2h}}\right) }\tag 4 $$
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c) We draw $\dfrac{1}{m}$ versus $R$ graph, as we see below and we found the slope of the best-fit line.
Now we need to plug this value into (4) to find the impulse of the spring.
$$ J_x= {0.24} \left(\sqrt{\dfrac{9.8}{2\times1.5}}\right) $$
$$ J_x= \color{red}{\bf0.434}\;\rm kg.m/s $$
