Aball is kicked horizontally off a cliff with an initial speed of 6 m/s and lands 12 m from the base of the cliff. for
how many seconds was the ball in the air? use a=9.8m/s2 for acceleration due to gravity.(enter only the number)

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                                

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.