Respuesta :
Answer
The time for ball in the air is, t = 2 s
Explanation:
Given,
The initial speed of ball, v = 6 m/s
The ball lands at a distance from the cliff are, S = 12 m
Since the ball is launched from a height 'h' horizontally with an angle θ = 0
Therefore, the initial vertical component of velocity Vy = 0
The initial horizontal component of velocity, Vx = 6 m/s
To find the height of the cliff, let us use the range formula
            S = Vx · [Vy + √(Vy² + 2gh) ] / g    meter
Since Vy = 0; squaring both sides and solving for h
            g²S² = Vx² + Vy² + 2gh
              h = S²g / 2Vx²    meter
Substituting the given values in the above equation
               h =  (12² x 9.8) / (2 x 6²)
                = 19.6 m
Therefore the height of the cliff is 19.6 m
To find the time of flight, the formula h > 0 is
               t = [Vy + √(Vy² + 2gh) ] / g    second
Since Vy = 0
               t = √(2h/g)   second
Substituting the values in the above equation
               t = √(2 x 19.6 / 9.8)
                = 2 s
Hence, the time for the ball in air is, t = 2 s                Â
