Recent content by Nicaragua

So is the total mass that has been given the velocity of 7.5 ms^-2 2.24 kg?

I am really sorry but I am re-visiting this question and do not understand it. Can you please explain to me how you work out the kinetic energy transferred when leaving the bottle? Do you use 1/2 m v^2 with v = the final velocity of the fluid (7.5 m/s^2)?

Wait, but the vertical velocity does change because of acceleration due to gravity?

Oh yes that's what I meant! Thank you so much for helping me this evening, we got there in the end :) I really appreciate it!

Ohhh I get it. SO the final velocity in the question was -5.745 m/s because it is a hyperbola and each point on the trajectory has a symmetrical opposite? And the velocity changes because v = u + at so as time increases, so does the final velocity?

So both horizontal and vertical velocities do not change during flight? Only the vertical one changes direction?

I'd say the total horizontal distance travelled, so you would use the horizontal velocity? But doesn't horizontal velocity change? Or is it the same throughout the flight?

So know I know the total flight time, so I can use the distance-time equation, distance = speed x time. Now the only problem is I am not sure which velocity to plug in, 7.5 m/s as the resultant, or 4.8 m/s as the horizontal velocity? Either way, the answers are 5.6 metres using the horizontal...

-9.81 m/s/s so using t = v-u / a, should it be t = -5.745 - 5.745 / -9.81 ? Which equals 1.171 seconds?

I don't understand what you mean. I just drew a sketch and know the initial horizontal and vertical velocities, and i pres-sume the final vertical velocity is -5.7 m/s. No wait, it should be 0 m/s? Ahhh this is so frustrating that you're not allowed to be clear with me.

Please can you take me through how you would solve the range question and I can ask any questions afterwards? I am just so confused now...with every reply I feel more and more lost. It would really help me see if you take me through the question. Thanks man

Oh ok so I would still use v = u + at? Just v = -5.7 m/s and u = 5.7 m/s ? Thanks for question b, I totally get it now.

So in terms of vertical components, does u=v? I might have been mixing it up with the horizontal thinking v=u. Anyway, what equation would even solve this? Ahh I see. So the liquid has a mass of 1.45kg. 1/2 x 1.45 x 7.5^2 = 40.781 Joules

So what quantities of the vertical component do I have? Acceleration, initial velocity and that's it? As for question b, it doesn't say how much water is left, only the total mass before and after. Can you please tell me how I go about this? I appreciate your help in trying to make me figure it...

For v I will put the value as 0 m/s because that's when the liquid will hit the ground/ be at its peak. For u I will put 6 m/s because this is what I found out previously. And for the kinetic energy, can you give me some more help because I am really struggling for ideas.