fountain height

[Ice-cream]

Can any1 help me with these 3 questions?

 

1. A fountain rises to a height of 560ft. a) what is the initial speed of the water? b) how long does it take for the water to raeach the top of the fountain? (ans:58m/s, 5.9s)

 

I used v = u + 2as where v = 0, u =? and s = 17.07m (after converting 560ft to cm then to m...1ft = 30.48cm right?) I found u to be 18.29m/s. I don't understand why my answer is so way off!

 

2. Wrongly called for a foul, an angry basketball player throws the ball straight down to the floor. If the ball bounces straight up and returns to the floor 2.5s after first striking it, what was the ball's greatest height above the floor?

 

I can't seem to gather enough info to do anything!!! I don't know initial or final velocity....what can i do?? (BTW, there wasn't an answer in the textbook for this...)

 

thanx

[swansont]

Can any1 help me with these 3 questions?

 

1. A fountain rises to a height of 560ft. a) what is the initial speed of the water? b) how long does it take for the water to raeach the top of the fountain? (ans:58m/s' date=' 5.9s)

 

I used v = u + 2as where v = 0, u =? and s = 17.07m (after converting 560ft to cm then to m...1ft = 30.48cm right?) I found u to be 18.29m/s. I don't understand why my answer is so way off!

 

2. Wrongly called for a foul, an angry basketball player throws the ball straight down to the floor. If the ball bounces straight up and returns to the floor 2.5s after first striking it, what was the ball's greatest height above the floor?

 

I can't seem to gather enough info to do anything!!! I don't know initial or final velocity....what can i do?? (BTW, there wasn't an answer in the textbook for this...)

 

thanx[/quote']

 

560 ft = 170.7 m

 

A little "reality check" of multiplying 17 X 3 would have told you something was wrong.

 

s = vt + 1/2 at² might be useful for the second part, and for problem 2.

 

for #2, use the symmetry of the problem - if there's no air resistance, the up travel and down travel take the same amount of time.