A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance.

Strategy
Draw a sketch.

We are asked to determine the position at various times. It is reasonable to take the initial position y₀ to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it. Since we are asked for values of position and velocity at three times, we will refer to these as y₁ and v₁; y₂ and v₂ ; and y₃ and v₃.

Transcribed Image Text:

20=13.0 m/s J L a = -9.8 m/s² X