#61
as we know that
[tex]tan\theta = \frac{height}{base}[/tex]
[tex]tan30 = \frac{AK}{100}[/tex]
[tex]AK = 100 tan30[/tex]
[tex]AK = 57.7 m[/tex]
#62
initial speed of the sphere = 0 as it is dropped
initial height of the ball = AK = 57.7 m
now by calculations
[tex]d = v_i* t + \frac{1}{2}at^2[/tex]
[tex]57.7 = 0 + \frac{1}{2}*9.8*t^2[/tex]
[tex]57.7 = 4.9t^2[/tex]
[tex]t = 3.43 s[/tex]
so it will take 3.43 s to reach the ground