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October 1, 2014

October 1, 2014

Posted by **some1** on Wednesday, February 6, 2008 at 7:16pm.

(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?

- Physics -
**Damon**, Wednesday, February 6, 2008 at 7:20pmIf 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 7:22pmwhat is U...meaning acceleration?....

so basically i would have

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

- Physics -
**Damon**, Wednesday, February 6, 2008 at 7:31pmNow 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 7:44pmwhoops, 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 7:49pmI 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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