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
