Recent content by Shiz Stain

ah ok thanks for the help so when the ramp is at 0.25m and the golf ball is released and goes down the ramp it will land 1 m away from the table. Thanks for the help really appreciate it and next time will try to look at a problem in different ways, but the good news is you helped me on how to...

Oh ok will do, just one question I now treat this as a trig equation and do this sin(A) = opposite / hypotenuse sin (0.25) = X / 2.15 m 0.25 * 2.15 = x 0.53 m =x x = 53cm or 0.53 m

Ah ok this is what I did a =g * sin x x = a * d / g x = 1.13 * 2.15 / 9.8 x = 1.13 * 0.22 x = 0.25

so the angle is 10 or 0.2?

Um, could you tell me the formula to find the angle? I seemed to have forgot that one

So the acceleration I found was correct? So far i got this g = 9.8 or 10 m/s² t = 0.45 s vf = 2.2 m/s a = 1.13 m/s² So what your telling me now is that I have to find the angle of the ramp?

Oh man, this is really making me feel slow sorry for the trouble vf² = Vi² + 2a (2.15m) 2.2² = 0 + 2 a (2.15 m) 4.84 = 4.3 a a = 1.13 m/s²

Ah man, i was thinking about using it but every time I did you said it was wrong haha, vf² = Vi² + 2a (2.15m) 4.3 = 0 + 4.3 a 18.5 = 4.3 a a = 4.3 m/s²But i think the vf i used is wrong i think it should be something like this vf² = Vi² + 2a (displacement) vf² = 0 + 2 (9.8) (2.15m) vf² = 42.14...

So its doing down the ramp I just the distance of the ramp? displacement = Vi t + 1/2 a t² 2.15 m = 0 + 0.5 a 2.2 ² 2.15 m = 0.5 *4.84 a 2.15 m = 2.42 a a = 0.9 m/s

how fast it must be going I guess you mean the speed? So its s = d/t s = 1 m / 0.45 s = 2.2 m/s Is that the answer to your question now?

So I am guessing the horizontal velocity is the same as Vf vf² = vi² + 2a displacement vf ²= 0 + 2 (10) (1.02) vf ²= 20 (1.02) vf² = 20.4 vf = 4.5 m/s Is that it? Or is it something else reason I am using 1.02m is because I added the 1m + 2cm the golf ball travels on the table if I just use...

well the thing is, when it golf ball goes down the ramp and travels 2cm across the table and then falls, it has to land exactly 1m away from the table. So what do I have to find then now? Do i just use all the information I have and find the x ( the height for the ramp) ? Or is there something...

What do you mean what velocity gives me 1m covered in 0.45s? Is it final velocity? If so then i just have to plug it in and solve for Vf?

Ah ok I think I know what you are trying to say You want me to find the acceleration then? If so here it is a = delta v / delta t a = 4.3 - 0 / 0.45 a = 4.3 / 0.45 a = 9.5 m/s ² If not could explain it in simpler terms or tell me what I need to actually do Thanks

Ah sorry it took a while for me to reply, and thanks for the tips so far I got this far Vf² = Vi² + 2a (displacement) Vf² = 0 + 2 (10 or 9.8) (0.92) Vf² = 20 (0.92) Vf² = 18.4 Vf = 4.3m/s Now I am thinking I am suppose to find time so I plan to use this equation: displacement = Vi (t) + 1/2 a...