You throw a ball straight up from the edge of a cliff. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected. Take the positive y-direction to be upward, and choose y = 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 = 3.159 m

Explanation:

Using the second equation of motion.

s = ut + 0.5at^2 1

Since the ball is thrown upward the acceleration due to gravity act against it.

Given:

distance s = y

Initial speed u = 12.0 m/s

Time t = 0.30 s

Acceleration a = -g = -9.8m/s^2

Equation 1 becomes;

y = ut - 0.5at^2

Substituting the given values. We have;

y = 12.0(0.30) - 0.5(9.8 × 0.3^2)

y = 3.6 - 0.441

y = 3.159 m

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

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