PH01 Kinematics Question....

Can't comment on the exam as I'm not taking it.

Regarding the question, I presume you're familiar with "suvat", or some permutation of that acronym.


I'd use s=ut+1/2 at^2 as your basic equation.

Form an equation for how far it travels upto the final second.

And form an equation as to how far it falls in total.

Then elminate s between the two equations to get t (yes, it's a quadratic).

Etc.

Post working if you're still stuck

Ok that is helpful but I am having trouble solving it as there are 3 unknowns: s, t and u (in the 2nd equation).

For the first part I got s=0t + 1/2 x 10 x t^2
s=0t + 5t^2

And the second part s=ux1 + 1/2 x 10 x 1^2
s=u+5


Any thoughts?

Your second equation should be for how far it falls in TOTAL, not just the last second; so u is again zero.

Take care with s and t in your second equation as from the first equation you have s is half the total distance, and t is the time upto half the distance; so your values for s and t in the second equation need to be considered with care.

Could you show me how you would get the answer? I'm having real trouble here. By the way it's not homework or anything, I just want to understand how it's done. Cheers.

Since in your first equation you've used s to respresent the distance traveled up to the last second, and this is half the total distance, then the total distance travelled, including the last second is 2s.

Since the time up to the last second is "t" from your first equation, then the total time including the last second is t+1.

So your second equation is 2s = 0t + 1/2 x 10 x (t+1)^2

i.e. 2s = 5 (t+1)^2.

Now eliminate s between your two equations to get a quadratic in t. Only one solution is valid.

Then work out how far it travels in the time t+1 to get the height it was dropped from.

Ok, I think I've got it...

t=2.4 so the total time worked out as 3.4, and therefore the total distance is about 58.28m

Cheers for all your help

That's some serious rounding you've done there. I make it $t=1+\sqrt{2}$