# Finding acceleration and distance

I think i found the answers but the solutions dont make sense to me. Please help me correct my solution. Thanks!

A soccer ball is released from rest at the top of a grassy incline. After 6.3 seconds, the ball travels 51 meters. One second later, the ball reaches the bottom of the incline.
a) What was the ball's acceleration? (Assume that the acceleration was constant.)
b) How long was the incline?

a)
at rest t=0 and yi=0
at t=6.3 yf=51 m
yf-yi= vi(t)+1/2(a)(t)^2
51m=0(6.3)+1/2(a)(6.3 s)^2
[(51m)(2)]/39.69s^2=a=2.57 m/s^2

b)
a(constant)=2.57m/s^2
t(final)=7.3s

deltay=vi(7.3s)+1/2(a)(t)^2
deltay=0+1/2(2.57m/s^2)(7.3s)^2
deltay=68.5m

## Answers and Replies

berkeman
Mentor
I think i found the answers but the solutions dont make sense to me. Please help me correct my solution. Thanks!

A soccer ball is released from rest at the top of a grassy incline. After 6.3 seconds, the ball travels 51 meters. One second later, the ball reaches the bottom of the incline.
a) What was the ball's acceleration? (Assume that the acceleration was constant.)
b) How long was the incline?

a)
at rest t=0 and yi=0
at t=6.3 yf=51 m
yf-yi= vi(t)+1/2(a)(t)^2
51m=0(6.3)+1/2(a)(6.3 s)^2
[(51m)(2)]/39.69s^2=a=2.57 m/s^2

b)
a(constant)=2.57m/s^2
t(final)=7.3s

deltay=vi(7.3s)+1/2(a)(t)^2
deltay=0+1/2(2.57m/s^2)(7.3s)^2
deltay=68.5m

Looks fine to me. Why do you think it doesn't make sense?

This is my first physics class (university physics) and I always feel like I'm missing something.. I like to have confirmation from someone that knows what they r doing ;-)

Can you give me a hint on question c please? Thanks!!

berkeman
Mentor
This is my first physics class (university physics) and I always feel like I'm missing something.. I like to have confirmation from someone that knows what they r doing ;-)

Can you give me a hint on question c please? Thanks!!

Did you post a c) question? Did you post your attempt at a solution for whatever c) is?

Question c was in a different post.. I'm sorry!! I made a mistake..