A ball is thrown vertically upward from the top of a cliff with a height of [tex]111 \, \text{m}[/tex] and an initial speed of [tex]38.8 \, \text{m/s}[/tex]. How long will it be in the air (in seconds)? (Assume [tex]g = 9.8 \, \text{m/s}^2[/tex])

A ball is thrown vertically upward from the top of the cliff of height x = 111 m with the speed of 38.8 m/s. The ball will be in the air for approximately 4.76 seconds.

To find out how long a ball thrown vertically upward from the top of the cliff of height x = 111 m with the speed of 38.8 m/s will be in the air, we can use the following formula: s = ut + 1/2 at²

Where: s = displacement u = initial velocity t = time t = √(2s/g) Where: g = acceleration due to gravity

We know that the initial velocity of the ball is 38.8 m/s, and the height of the cliff is 111 m.

Thus, the initial displacement of the ball is 111 m above the ground. The acceleration due to gravity is given as 9.8 m/s².Using the formula,t = √(2s/g)t = √(2 × 111 / 9.8)t = √(22.65)t = 4.76 seconds (rounded to two decimal places)

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