Recent content by Samyuerux

Ok so far i have velocity of balls before hitting the ground using d = 1/2at^2 + Vit this works out as 2= 5t^2 therefor t = 0.28 seconds And then using V = a.t i get V at 2.8 seconds (g = 10m/s^2) for this example Then for velocity after collision i find the Ep (potential energy) = m.g.h = 200...

Just realized they will both have same velocity.. lol and velocity will be about 2.8 m/s i believe

Thanks very much guys got their in the end! :)

Homework Statement 4. The double-ball bounce Consider 2 balls on top of another that are being dropped from a height h = 2 m, see the Figure below. They will hit the ground and bounce back and we want to compute the velocity of the upper ball after the bounce. (a) (b) We can model this...

so then i rearrange to get 12 tan 30 = t so t = 6.93 and then times t by Vx = 83.16 and t by Vy = 13.86 then do pythag to find the total displacement? which comes out at 84.3 that seems too high...

I think i get you now, so far i have Sy = t^2 and Sx = 12t so i think i set Sy = t^2 as xtan30 = t^2

Well i set up a basic right angled triangle with the Hyp being the slope downwards at 30 degrees. The horizontal would then be set at 12. this gives me a vertical componant that's 6.93, so i presumed that would be the distance but i think that's wrong and there is a more complicated method of...

is it 6.93 Metres? Fast reply would be great thank you!

IM not sure how you work it out ?

I am having difficulties with this question as no height for the crater is given please help! :) Homework Statement AN astronaut is speeding over a level stretch of the Moon's surface at 12m/s in her small lunar rover. She drives over the edge of a crater, whose surface slopes downwards at an...