Question

While standing at the edge of a cliff, you throw a stone upward with an initial speed of 6.79 m/s. The stone subsequently falls to the ground, which 12.1 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? Ignore air resistance and take g= 9.81 m/s^2 a) 13.6 m/s

b) 14.5 m/s
c) 15.3 m/s
d) 16.8 m/s

Answer 1

Answer: d) 16.8 m/s

Step-by-step explanation:

In this situation we are dealing with paarabolic motion with a constant acceleration (acceleration due gravity), hence, the following equation will be useful to find the answer:

`{V_(f)}^(2)={V_(o)}^(2)+2ad`

Where:

`V_(f)` Is the final velocity of the stone

`V_(o)=6.79 m/s` Is the initial velocity of the stone

`a= 9.8 m/s^(2)` is the constant acceleration of the stone due gravity

`d=12.1 m` is the height of the cliff

`V_(f)=\sqrt{{V_(o)}^(2)+2ad}`

`V_(f)=\sqrt{{(6.79 m/s)}^(2)+2(6.79 m/s)(12.1 m)}`

`V_(f)=16.8 m/s` This is the final velocity of the stone