Conquer Club

A golf ball rolls off a horizontal cliff with an initial speed of 12.3 m/s. The ball falls a vertical distance of 18.7 m into a lake below. How much time does the ball spend in the air?

unfortunately i was absent for the first time this year on the day we did stuff like this...

ParadiceCity9 wrote:A golf ball rolls off a horizontal cliff with an initial speed of 12.3 m/s. The ball falls a vertical distance of 18.7 m into a lake below. How much time does the ball spend in the air?

unfortunately i was absent for the first time this year on the day we did stuff like this...

Well, the initial speed doesn't matter since it's all horizontal, so just calculate how long it takes for a ball to drop 18.7 meters. I haven't done a problem like this in a while so I forget the exact formula.

Frigidus wrote:

ParadiceCity9 wrote:A golf ball rolls off a horizontal cliff with an initial speed of 12.3 m/s. The ball falls a vertical distance of 18.7 m into a lake below. How much time does the ball spend in the air?

unfortunately i was absent for the first time this year on the day we did stuff like this...

Well, the initial speed doesn't matter since it's all horizontal, so just calculate how long it takes for a ball to drop 18.7 meters. I haven't done a problem like this in a while so I forget the exact formula.

Yes, in this problem it would not matter. So since gravity pulls at a rate of 9.8 m/s, I would think it is simply just:

12.3/9.8 m/s

The Neon Peon wrote:

Frigidus wrote:

ParadiceCity9 wrote:A golf ball rolls off a horizontal cliff with an initial speed of 12.3 m/s. The ball falls a vertical distance of 18.7 m into a lake below. How much time does the ball spend in the air?

unfortunately i was absent for the first time this year on the day we did stuff like this...

Well, the initial speed doesn't matter since it's all horizontal, so just calculate how long it takes for a ball to drop 18.7 meters. I haven't done a problem like this in a while so I forget the exact formula.

Yes, in this problem it would not matter. So since gravity pulls at a rate of 9.8 m/s, I would think it is simply just:

12.3/9.8 m/s

You mean 18.7m/9.8m/s...

Frigidus wrote:

ParadiceCity9 wrote:A golf ball rolls off a horizontal cliff with an initial speed of 12.3 m/s. The ball falls a vertical distance of 18.7 m into a lake below. How much time does the ball spend in the air?

unfortunately i was absent for the first time this year on the day we did stuff like this...

Well, the initial speed doesn't matter since it's all horizontal, so just calculate how long it takes for a ball to drop 18.7 meters. I haven't done a problem like this in a while so I forget the exact formula.

I'm using the equation x=x(naught) + v(naught)t -1/2at^2

..so i'm pretty sure initially velocity matters.

i have the equation:

-4.9t^2 + 12.3t - 18.7 = 0

is that right? then just solve for t..

hecter wrote:

The Neon Peon wrote:

Frigidus wrote:

ParadiceCity9 wrote:A golf ball rolls off a horizontal cliff with an initial speed of 12.3 m/s. The ball falls a vertical distance of 18.7 m into a lake below. How much time does the ball spend in the air?

unfortunately i was absent for the first time this year on the day we did stuff like this...

Well, the initial speed doesn't matter since it's all horizontal, so just calculate how long it takes for a ball to drop 18.7 meters. I haven't done a problem like this in a while so I forget the exact formula.

Yes, in this problem it would not matter. So since gravity pulls at a rate of 9.8 m/s, I would think it is simply just:

12.3/9.8 m/s

You mean 18.7m/9.8m/s...

No i did that with the corresponding example in the book and it came out different than the answer. If it helps, the question in the book says that the initial velocity is 11.4 and it goes down 15.5m and the answer there is 1.78 s.

Horizontal speed should only matter for distance travelled, not for time spent in the air, as gravity is constant. The moment it doesn't have the force of the cliff pushing up on it, then the ball will fall at 9.8m/s.

hecter wrote:Horizontal speed should only matter for distance travelled, not for time spent in the air, as gravity is constant. The moment it doesn't have the force of the cliff pushing up on it, then the ball will fall at 9.8m/s.

ya i know but it's not the right answer for some reason.

just thought of using parametric equations...the one i used before and x=12.3t?

Well, my advice to you is don't do it and ask the teacher tomorrow.

ParadiceCity9 wrote:just thought of using parametric equations...the one i used before and x=12.3t?

hrm just graphed that and it's all in quadrant IV..not good.

wouldn't x=12.3t be a graph for a constant velocity and therefore would not account for acceleration due to gravity?

strike wolf wrote:wouldn't x=12.3t be a graph for a constant velocity and therefore would not account for acceleration due to gravity?

ya but it would give the horizontal motion of the ball.

umm, did it roll off at an angle? that would mean there was an initial X velocity

Herakilla wrote:umm, did it roll off at an angle? that would mean there was an initial X velocity

no indication of an angle. initial x velocity is definitely 12.3 though i know that...

The Neon Peon wrote:

Frigidus wrote:

ParadiceCity9 wrote:A golf ball rolls off a horizontal cliff with an initial speed of 12.3 m/s. The ball falls a vertical distance of 18.7 m into a lake below. How much time does the ball spend in the air?

unfortunately i was absent for the first time this year on the day we did stuff like this...

Well, the initial speed doesn't matter since it's all horizontal, so just calculate how long it takes for a ball to drop 18.7 meters. I haven't done a problem like this in a while so I forget the exact formula.

Yes, in this problem it would not matter. So since gravity pulls at a rate of 9.8 m/s, I would think it is simply just:

12.3/9.8 m/s

Ah, ah, but that's just how fast the ball accelerates per second. Since it isn't a constant speed it isn't that easy. Basically -9.8 would be the change in slope. This treads into calculus territory, something I managed to wash my mind of after my freshman year of college.

ParadiceCity9 wrote:
-4.9t^2 + 12.3t - 18.7 = 0

If someone can solve that or tell me how to solve it on the TI-89 Titanium then that's all I need.

ParadiceCity9 wrote:

Herakilla wrote:umm, did it roll off at an angle? that would mean there was an initial X velocity

no indication of an angle. initial x velocity is definitely 12.3 though i know that...

Horizontal is an indication that there is no angle to affect downward velocity.

strike wolf wrote:

ParadiceCity9 wrote:

Herakilla wrote:umm, did it roll off at an angle? that would mean there was an initial X velocity

no indication of an angle. initial x velocity is definitely 12.3 though i know that...

Horizontal is an indication that there is no angle to affect downward velocity.

...and that's what i just said.

blah i just remembered how to do it, you guys are where my class started like 6 weeks ago

you had the right equation but the initial velocity is 0 so its 18.7 = T(0) + (1/2)(9.8)(T^2)

answer is 1.95 secs depending on how you round

Herakilla wrote:blah i just remembered how to do it, you guys are where my class started like 6 weeks ago

you had the right equation but the initial velocity is 0 so its 18.7 = T(0) + (1/2)(9.(T^2)

answer is 1.95 secs depending on how you round

we always round to 3 sig figs. thanks man.

new question
i've got 6 more, progressively getting harder so i will probably ask for help on most..


A golf ball rolls off a horizontal cliff with an intial speed of 11 m/s. The ball falls a vertical distance of 15.5 m into a lake below. What is the speed v of the ball just before it strikes the water?

actually i think i know how to do this...find the amount of time it's in the air (t) and multiply that by -9.8 then add that to 11?

er..is it not negative 9.8?

well youre only talking about one direction so negative doesnt matter, the ball doesnt go up and then down, it only goes down. hold on a second and ill post you an easier way