You throw a ball straight up from the edge of a cliff. It leaves your hand moving at 12.0 m/sm/s. Air resistance can be neglected. Take the positive yy -direction to be upward, and choose yy = 0 to be the point where the ball leaves your hand. Find the ball's position 0.300 s after it leaves your hand.

y = 3.16 m

Explanation:

Once thrown straight up, the only influence on the ball is gravity, which acts slowing down the ball.

In absence of gravity, the ball had moved forever  with the same initial velocity, so the displacement at a time t had been as follows:

y = v₀*t

Due to the action of gravity, the actual displacement is less than this value, and can be calculated with the following equation:

Replacing by v₀ = 12. 0 m/s, t = 0.300 s, and  g = 9.8 m/s², we  can solve for the vertical displacement, as follows:

So, the position (taking the origin coincident with the point where the ball leaves the hand), is just the vertical displacement we have just found.

⇒ y = 3.16 m

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .