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.)

Question

(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
x-x(0)= V(0)xt + 1/2at^2
y-y(0) = V(0)yt - 1/2at^2
but i dunno if i am using the right equation....can someone help me plz?

Answer 1

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

Answer 2

what is U...meaning acceleration?....

so basically i would have

1.80= -9.8t?...is that right?

Answer 3

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

Answer 4

whoops, I used 1.90 instead of 1.80 for table height. You will have to redo the arithmetic :)

Answer 5

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