I have another quick question. It looks easy, but for some reason I just don't think I'm doing it right.

Please check over this:

"A kid throws a ball straight up into the air. The ball is caught 2.5 seconds later. "

a) How high did the ball go (from its point of release)?
b) What was the speed of the ball the moment before it was caught?





The first thing that threw me off was no initial velocity was given... I assume it's 9.8 m/s?

My solutions:

a) x = .5(vi+vf)(t)
x = .5(9.8)(2.5/2) = 6.125 m

I divided the time since you can conside both the up and down direction symmetrical.

I believe that is correct, but not sure about this b part:

b) vf = vi + at
vf = 9.8(2.5/2) = 12.25

What do you do here when there is no initial velocity? Something just doesn't make since about this problem.





The next question is this:

"A rock is thrown downward from a bridge with a speed of 15 m/s. If the bridge is 50m above a river, how long will it take for the rock to hit the water?"

My solution:

x = vi(t) + .5a(t)^2

-50 = 15t + .5(-9.8)t^2

Solve as quadratic (after multiplying by negative one):

4.9t^2-15t-50 = 0

formula:

t = 15 +- sqrt(15^2 - 4(4.9)(-50) )
--------------------------------------
9.8

and you get: 5.07 seconds... Is that correct?




Any help/clarity would be much appreciated.


PS: These aren't homework questions. My teacher doesn't give homework and that means I have to do some serious studying to prepare for tests.