Answer
$475\;\rm m/s^2$
Work Step by Step
We can divide the trip of the javelin into two stages:
1) When it was attached to the athlete's hand and accelerated at a constant rate.
2) When it moves in the air under free-fall acceleration.
1) The javelin was accelerated from rest at a constant acceleration for a distance of $d=0.7$ cm.
Thus,
$$v_{1}^2=\overbrace{v_{i}^2}^{0}+2ad$$
whereas $v_{1}$ is the releasing speed when the athlete lets the javelin fly away.
Solving for $a_x$;
$$a =\dfrac{v_{1 }^2}{2d}\tag 1$$
2) Now we need to find $v_{1 }$ which is the initial velocity of the javelin in its air trip.
We know that it falls after 62 m from the throwing point.
$$x_f-x_i=v_{ix}t_2$$
$$62-0=v_{1}\cos 30^\circ t_2$$
whereas $t_2$ is the time of the air trip.
Thus,
$$t_2=\dfrac{62}{v_{1}\cos 30^\circ }\tag 2$$
Now we need to use the kinematic formula of vertical displacement.
$$\overbrace{y_f}^{0}=y_i+v_{iy}t_2-\frac{1}{2}gt_2^2$$
$$0=2+v_{1}\sin 30^\circ t_2-4.9t_2^2$$
Plugging from(2);
$$0=2+v_{1}\sin 30^\circ \left[\dfrac{62}{v_{1}\cos 30^\circ }\right]-4.9\left[\dfrac{62}{v_{1}\cos 30^\circ }\right]^2$$
$$0=2+62\tan30^\circ-\dfrac{4.9\cdot 62^2}{v_1^2\cos^230^\circ}$$
$$ \dfrac{4.9\cdot 62^2}{v_1^2\cos^230^\circ}= 2+62\tan30^\circ$$
$$ v_1^2=\dfrac{4.9\cdot 62^2}{ [2+62\tan30^\circ]\cos^230^\circ} $$
$$ v_1 =\sqrt{\dfrac{4.9\cdot 62^2}{ [2+62\tan30^\circ]\cos^230^\circ} }$$
$$v_1=25.78\;\rm m/s\tag {Plug into (1)}$$
$$a =\dfrac{25.78^2}{2d}=\dfrac{25.78^2}{2\cdot 0.7}$$
$$a =\color{red}{\bf 475}\;\rm m/s^2$$
