Question

Projectile motion and time of flight and maximum height

Answer 1

Time of flight = 2u sinФ/g
Maximum height = (u sinФ)^2 / 2g

Answer 2

Total Time of flight = T =2 *  (u sin Ф) / g

 H = (ut -1/2 a t²  formula)
   = u Sin Ф * (u sin Ф / g) - 1/2 g (u SinФ/g)²
   =  u² Sin² Ф / 2 g

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I am deriving the formulas for you here in detailed manner.

    Let us say the the projectile is fired into air at an angle of Ф with the horizontal and with an initial speed of u in that direction.  Initially, at time t = 0sec, the projectile is at origin O(0,0) at height =0.  At time t sec, the projectile is on the parabolic path at a point P (x, y) with instantaneous velocities Vx, and Vy in x and y directions.

There is a deceleration in the vertical direction = - g

Speed of projectile in the vertical direction is : 
   V = u + a * t    =>  Vy = u sin Ф - g t        --  equation 1

Speed in the horizontal direction is a constant as there is no acceleration:
       =>    Vx = u cos Ф + 0 * t      --  equation 2

Displacement = Distance traveled is :  s = u t + 1/2 a t²  
        =>    Sx = x = (u cos Ф) t         --- equation  3
                   t = x / (u cos Ф)

            Sy = y = (u Sin Ф)  t  -  1/2 g t²
            y =  (u sin Ф) (x / u cos Ф) - 1/2 g (x² / u² Cos² Ф)
           y = x tan Ф - g x² / (2u²Cos²Ф)           ---   equation 4

This is the equation of motion of a 2-d projectile and its locus.

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The maximum height H is reached by projectile when  Vy becomes 0.
=>  Vy = 0 = u sin Ф - g t   

=>  t  =  u sin Ф / g  = time interval to reach the maximum height.
=>  H = u Sin Ф * (u sin Ф / g)  - 1/2 g (u SinФ/g)²
=>  H =  u² Sin² Ф / 2 g    --   equation  5

Time of flight = 2 * (u sin Ф) / g          --- equation 6

  y = 0 =>  x = R = Range of the projectile , where it lands on ground.
Substituting in equation 4, we get
            R = u² Sin 2Ф / g       --  equation 6