Posted by some1 on Wednesday, February 6, 2008 at 7:16pm.
A ball rolls horizontally off the edge of a tabletop that is 1.80 m high. It strikes the floor at a point 1.42 m horizontally away from the table edge. (Neglect air resistance.)
(a) How long was the ball in the air?
(b) What was its speed at the instant it left the table?
i have the equation
xx(0)= V(0)xt + 1/2at^2
yy(0) = V(0)yt  1/2at^2
but i dunno if i am using the right equation....can someone help me plz?

Physics  Damon, Wednesday, February 6, 2008 at 7:20pm
If x is horizontal, no you are not using the right equation.
Gravity has no horizontal component
You horizontal speed, call it U is constant
so
X  Xo = U t
In the vertical direction, fine with a = g = 9.8 m/s

Physics  some1, Wednesday, February 6, 2008 at 7:22pm
what is U...meaning acceleration?....
so basically i would have
1.80= 9.8t?...is that right?

Physics  Damon, Wednesday, February 6, 2008 at 7:31pm
Now to continue
call the initial X0 = 0
the final X = 1.42 m
so
1.42 = U t
where U is that initial speed which remains the horizontal speed and t is time in the air
NOW vertical direction
call y upwards from the top of the table
so Yo = 0
Yfinal =  1.90 m , the floor
Vo = 0, no vertical speed originally, only the horizontal speed U
so
1.90 = 0  (1/2)(9.8) t^2
t^2 = 1.9/4.9 = /0.388
so
t = 0.623 seconds in the air
now back to get U
1.42 = U t = U (.623)
so
U = 2.28 m/s

Physics  Damon, Wednesday, February 6, 2008 at 7:44pm
whoops, I used 1.90 instead of 1.80 for table height. You will have to redo the arithmetic :)

Physics  Damon, Wednesday, February 6, 2008 at 7:49pm
I called U the horizonal speed, and V the vertical speed.
The point is that this is two problems.
1. a constant speed horizontal problem where distance = horizontal rate * time
2. a vertical problem where distance is initial vertical speed times time plus (1/2) a t^2
The two problems are connected by the time in the air, t
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