(1) The ball is in the air for <u>1.4 seconds.</u>
(2) The horizontal velocity of the ball as it rolls off the table is<u> 6.32 m/s.</u>
(3) The vertical velocity of the ball right before it hits the ground is <u>13.72 m/s.</u>
(4) The horizontal velocity of the ball right before it hits the ground is<u> 6.32 m/s.</u>
(5) The initial vertical velocity as soon as the ball comes of the cliff is <u>13.72 m/s.</u>
<h3>What is the time of motion of the ball?</h3>
The time of motion of the ball is calculated by applying the following equation.
t = √(2h/g)
where;
- h is the height of the cliff
- g is acceleration due to gravity
t = √(2h/g)
t = √(2 x 9.63 / 9.8)
t = 1.4 seconds
The horizontal velocity of the ball is calculated as follows;
v = d/t
where;
- d is the horizontal distance travelled by the ball = 8.85 m
v = 8.85 m / 1.4 s
v = 6.32 m/s
The vertical velocity of the ball before it hits the ground is calculated as;
vf = vi + gt
vf = 0 + 9.8 x 1.4
vf = 13.72 m/s
The horizontal velocity of the ball right before it hits the ground is calculated as;
the initial velocity of a projectile = final horizontal velocity
vxf = vxi = 6.32 m/s
The initial vertical velocity as soon as the ball comes off the cliff = final vertical velocity = 13.72 m/s
Learn more about horizontal velocity here: brainly.com/question/24949996
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