Projectile Motion Calculations: Time in Air and Speed at Launch

[Kruz87]

Q: A small ball rolls horizontally off the edge of a tabletop that is 1.23 m high. It strikes the floor at a point 1.96 m horizontally away from the edge of the table. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

So this is what I did, i treated the vertical and horizontal motions separately. And for (a) I solved for the time by using the constant acceleration equation, with y-yo=-1.23, a=-9.8 m/s^2 and Voy=0. Using those values I got t= .50s. I then analyzed the horizontal motion, but I'm not understanding how to get the speed at the instant it leaves the table since at t=.50s would be the speed at which it hits the ground, how do I get the time at the instant it leaves the table in order to use the constant acceleration formula? Any help at all is appreciated, thanks

 

[Doc Al]

So this is what I did, i treated the vertical and horizontal motions separately. And for (a) I solved for the time by using the constant acceleration equation, with y-yo=-1.23, a=-9.8 m/s^2 and Voy=0. Using those values I got t= .50s.

Good.

I then analyzed the horizontal motion, but I'm not understanding how to get the speed at the instant it leaves the table since at t=.50s would be the speed at which it hits the ground, how do I get the time at the instant it leaves the table in order to use the constant acceleration formula?

How does the horizontal component of velocity change with time?

 

[Kruz87]

the horizontal component of velocity is constant.

 

[Kruz87]

Does that mean that I can take the velocity at any time t?

 

[Kruz87]

Or rather time t=.5s using x-x0=1.96 m

 

[Kruz87]

Yep, that was right! Thanks a lot Doc Al.

 

[Doc Al]

the horizontal component of velocity is constant.

Right!

Does that mean that I can take the velocity at any time t?

I'm not sure what you mean, since it's a constant. How do you propose to find the horizontal velocity?

 

[Doc Al]

Or rather time t=.5s using x-x0=1.96 m

Right. That will give you the horizontal velocity period, not just at t=.5s.

 

[Kruz87]

Ok, I see what you mean now, it just made since to use t=.5s because we had a horziontal distance for that value, Thanks again