you forgot a 2 in the first equation but then you put it back later so the answer should be right. How do you know it's not? (if you are using Webassign, it may be simply a question of sig figs)Spruance said:To solve for the initial speed v0:
v0^2 = v^2 + 2gh - gh0
v0 = (v^2 + 2gh - gh0)^1/2
v0 = (2g(h-h0))^1/2
v0 = (2 * 9.81 (4 m -3 m))^1/2
v0 = 4.43 m/s
What is wrong?
you shoudl have mentioned that you has switched to question 2 now...Spruance said:Ok.
Let us assume that the initial speed is 4.43 m/s
then the height of ball is given:
h = 1/2 (3/4 vo2)/g + ho =
3/8 (vo2)/g + ho
= 3/8 * ((4.43^2)/9.81) + 3
= 3.75 m
Is that correct?