Answer:
a)   v₀ = 32.64 m / s
, b)  x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
          x = v₀ₓ t
          y =  t - ½ g t²
 t - ½ g t²
We look for the components of speed with trigonometry
          sin 43 = v_{oy} / v₀
          cos 43 = v₀ₓ / v₀
          v_{oy} = v₀ sin 43
          v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
          t = x / v₀ cos 43
          y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
           y = x tan 43 - ½ g x² / v₀² cos² 43
           1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
            v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
           v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
           v₀ = √ (35280 / 33.11)
           v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
          y =  t - ½ g t²
 t - ½ g t²
          .y = v₀ sin43 t - ½ g t²
         25 = 32.64 sin 43 t - ½ 9.8 t²
         4.9 t² - 22.26 t + 25 = 0
          t² - 4.54 t + 5.10 = 0
We solve the second degree equation
          t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
          t = (4.54 ± 0.46) / 2
          t₁ = 2.50 s
          t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
          x = v₀ₓ t
          x = v₀ cos 43 t
          x = 32.64 cos 43  2.50
          x = 59.68 m