Answer:
(a) H = 1.41 m
(b)   H₁ + H₂ = 1.35 m
Explanation:
LONGER  TRACK  
To calculate the height H of the longer track, we use the equation of motion on an inclined plane:
V² = U² -2gH---------------------------------------------- (1)
H = (U²- V²)/ 2g------------------------------------------- (2)
Since the block came to rest at height H, it implies that the final velocity V =0
Vertical component of the Initial velocity U =   6.94Sin 50°
Substituting into (2)
H = (6.94Sin 50°)²/(20)
   = 1.4131
  = 1.41 m
SHORTER TRACK (First Motion)
For the shorter track, the velocity  (Vf)  of the block at the end of the track is calculated as thus:
Initial velocity , V₀=   6.94 m/s
The vertical component of the velocity is 6.94Sin 50°
From the Law of Equation:
V² = U² -2gH---------------------------------------------- (1)
Substituting into (1)
V²  =  (6.94 Sin 50⁰)² – (2 x10 x1.25)
     =  28.2635 – 25
     = 3.2635
  Vf    = √3.2635
         = 1.8065m/s
        =   1.81 m/s
SHORTER TRACK (2nd Motion)
The block flew off at the end of the track in a projectory motion as shown above. This implies that the velocity (Vf) will be tangential to the path of motion and inclined as 50⁰ to the horizontal.
 The vertical component of Vf   = 1.8065 Sin 50⁰  
 Initial Velocity U = 1.8065 Sin 50⁰  
At the maximum height of trajectory, final velocity, V = 0
To calculate H₂, we deploy the equation of motion in equation (1)
Substituting our new values into (1), we have:
0 = (1.8065 Sin 50⁰)² – (2 x10) x H₂
H₂ = (1.8065 Sin 50⁰)²/ 20
     =   0.09575 m
 H₁ + H₂ = 1.25 + 0.09575
              = 1.34575
              =  1.35 m