A ball is thrown vertically upward …

Using the relation, v² = u² + 2ax, we have
{tex}x = \frac { v ^ { 2 } - u ^ { 2 } } { 2 a } = \frac { 0 - ( 20 ) ^ { 2 } } { 2 \times ( - 10 ) }{/tex}= 20 m.
Hence, the ball will rise upto 20m and height from the ground = 20 + 25 = 45 m.

From the given figure, we have upward motion A to B in time(t₁) and downward motion(B to C) in time(t₂)
Also Velocity at B is zero.
v = u - gt
From the top, the ball falls freely under acceleration due to gravity
{tex}\therefore \quad x = x _ { 0 } + u t + \frac { 1 } { 2 } g t _ { 2 } ^ { 2 }{/tex}
Since x  = 0, x₀ = 25 m, u = 0 and g = 10 m/s²
Therefore, 25 + 20 {tex}\times t- \frac12 \times10 \times t^2=0{/tex}
{tex}\Rightarrow t^2-4t-5=0{/tex}
on solving, we get t = 5 and t = -1 
Time taken by the ball to hit the ground  = 5 s   (reject  t = -1 )