Need help with physics

[Jordan6969 0]

I got a big assignment due on monday and I have 2 questions that I totally do not know how to do. Any help would be appreciated. And as a side note the teacher gives us the answers to know if we are doing it right and ^2 means squared.

1. When an object is dropped from a height of 20.0m above the surface of planet Z, it will fall 5.00 m during the 2nd second of fall. What is the acceleration of a falling object near the surface of this planet. (The answer says 3.33. m/s^2)

2. A ball is rolled up a constant slope with an initial velocity of 11.0 m/s, and after 9.3 s the ball is roling down the slope with a velocity of 7.3 m/s. What is the rate of acceleration of the ball on the slope? (Answer says -2.0 m/s^2)

[peter54 0]

Alright, I think that I've got this right.

1. Given:

y(0)=5.0m

y(f)=0m

t=1s

Find:

a

Solution:

v=v(0)-a*t

y=y(0)+v(0)*t-[(1/2)*a*(t^2)]

v^2=v(0)^2-[2*a*{y-y(0)}]

substituting the given values into the equations give:

v=v(0)-a*1s

0=5.0m+v(0)*1s-[(1/2)*a*(1s^2)]

v^2=v(0)^2-[2*a*{5.0m-0m}]

-this gives three equations and three unknowns(v,v(0), and a)

-solving for "a" should give the answer of:

a=-3.33m/s^2

2. Given:

v(0)=11.0m/s

t=9.3s

v=-7.3m/s

Find:

a

Solution:

v=v(0)+a*t

-7.3m/s=11.0m/s +a*t

solving for "a" yields:

a=-1.96774m/s^2

rounded to two sig figs

a=-2.0m/s^2

I hope that helps!

[Jordan6969 0]

Thanks for your help peter, it took me a while to understand your work but at least I get it now.