Question

a student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 19.5 m/s. the cliff is 24.6 m above a flat, horizontal beach. x y 24.6 m 19.5 m/s g o how long after being released does the stone strike the beach below the cliff? the acceleration of gravity is 9.8 m/s 2 . answer in units of s. 013 (part 2 of 3) 10.0 points at impact, what is its speed? answer in units of m/s.

Answer 1

The time taken to reach the ocean from the cliff is 2.24 seconds

The speed of the stone at the point of contact is 21.984 m/s

The stone is thrown horizontally with a speed of 19.5 m/s from the cliff

The height of the cliff = 24.6m

• The time taken by the stone to reach the ocean can be calculated using the formula

                      y - v₁t = -1/2 gt²

where y is the height of the cliff

           v₁ is the speed of stone vertically

           g is the acceleration due to gravity

           t is the time taken

But the speed of the stone vertically will be zero, v₁ = 0

Therefore, the equation becomes

               y = -1/2gt²

Let us re-write this equation in order to find t

            t = √(-2y/g)

Now, let's substitute the known values in the above equation, we get
          t = √(-2x -24.6 / 9.8)

           =√(49.2/9.8)

           = √5.02

          = 2.24 seconds

Therefore, the time taken to reach the beach is 2.24 seconds

• The speed of the stone at the point of impact can be found using the formula

                                              a = v/t

where a is the acceleration

          v is the velocity

          t is the time taken

Here, the acceleration is the acceleration due to gravity

Thus, the velocity of the stone is

                    v = at

                      = 9.8 x 2.24

                     = 21.984 m/s

Therefore the velocity of the stone is 21.984 m/s

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