Answer with complete steps it took !

Y = -16.t² + 64.t+119.
It's a quadratic function where t = time and y = height
1) This object reaches the ground when y (height) = 0, hence let's calculate the roots (or the zero values) of t:
 -16t² + 64.t + 119 = 0 
t' =[-b + √(b²-4.a.c)]/2a       and t" = =[-b - √(b²-4.a.c)]/2a      

t'  = 6 sec    and t" = -2.05 (extraneous solution = inacceptable)

Ten time needed to reach the ground t = 6 seconds. 

2) This is a parabola open downward (a= -16<0), with a maximum that is equal to the vertex.
 
Axis of symmetry x = -b/2a = - (64)/(2)(-16) = -64/-32 = 64/32 = 2. Now to find y, plug in 2 in the equation:

-16(2²) + 64(2) + 190 = 254

Answers : Time to reach the ground → t = 2 sec
                 Maximum height→→→→→h = 254

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