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M1 Help?

Whilst doing revision I have come across 2 questions that I cannot seem to get the right answer for. Can anyone help?

1) A stone is trhown vertically upwards and takes 4 seconds to reach its highest point. Calulate the time in seconds to 2 significant figures, for which the stone is at least 45m above its point of projection. (answer from part a is u=39.2m/s)

The answer is - 5.2 s



2) A stone is dropped from a top of a cliff. One second later another stone is thrown vertically downwards from the same point with speed 16m/s. Both stones reach the bottom of the cliff at the same time wihtout hitting any obstructions. Find the height of the cliff.

The answer is - 16m


Thankyou :smile:
Reply 1
Avatar for ttoby
ttoby
15
1. We'll take upwards as positive. You need to use s=ut+12at2s=ut+\frac{1}{2} a t^2 where a=-9.8 and u=39.2 as you said. Make s=45 and then solve for t as a quadratic. This will give you the two times that the stone reaches exactly 45m, so for your final answer you take the difference of those two times.

2. We'll take downwards as positive. Using s=ut+12at2s=ut+\frac{1}{2} a t^2 again, we can say that h=0t+12gt2h=0t+\frac{1}{2} g t^2 for the 1st stone (where h is the height of the cliff and t is the time taken to fall).

Also, h=16(t−1)+12g(t−1)2h=16(t-1)+\frac{1}{2} g (t-1)^2 for the 2nd stone, since it starts at 16 m/s and it takes 1 s less time to fall.

Now you can solve these two equations simultaneously. First make them equal to each other to get t, then from there calculate h.
Reply 2
Avatar for YGB Jammy
YGB Jammy
1
First time using LaTeX markup so sorry if it's not upto scratch, also ask me if you need anything explained a bit clearer.

1)
u=39.2u=39.2
s=45s=45
a=−9.8a=-9.8
v=?v=?

v2v^2 = u2u^2 + 2as2as

v=25.59ms−1v = 25.59ms^{-1}

Then,
u=25.59u=25.59
v=0v=0
a=−9.8a=-9.8
t=?t=?

v=u+atv=u+at
t=2.6st=2.6s

Then x2 as the stone will fall back down again symmetrically.
t=5.22st=5.22s


2)
Let things to do with the first stone be labelled as 1, and things to do with the second be labelled as 2. Also let the height of the cliff = h.

Stone 1

u1=0u_1 = 0
a=9.8a = 9.8
s=hs = h
t1=t1t_1 = t_1

Stone 2

u2=16u_2 = 16
a=9.8a=9.8
s=hs=h
t2=t1−1t_2 = t_1 - 1

Now we can apply s=ut+0.5at2s=ut + 0.5at^2 to both equations to give us two equations:

Stone 1:

h=0+0.5×t12h = 0 + 0.5 \times t_1 ^2
h=4.9t12h = 4.9t_1 ^2

Stone 2:

h=16(t1−1)+0.5×9.8(t1−1)2h = 16 (t_1 - 1) + 0.5 \times 9.8 (t_1 -1)^2
h=16t1−16+4.9(t12−2t1+1)h = 16t_1 - 16 + 4.9 (t_1^2 - 2t_1 + 1)
h=16t1−16+4.9t12−9.8t1+4.9h = 16t_1 - 16 + 4.9t_1^2 - 9.8t_1 + 4.9
h=4.9t12+6.2t1−11.1h = 4.9t_1^2 + 6.2t_1 - 11.1

We now have two simultaneous equations defining h, so we can combine them:

4.9t12=4.9t12+6.2t1−11.14.9t_1 ^2 = 4.9t_1^2 + 6.2t_1 - 11.1
t1=1.79st_1 = 1.79s


Then using: h=4.9t12h = 4.9t_1^2
h=4.9×(1.792)h = 4.9 \times (1.79^2)
h=15.7mh = 15.7m


Really hope this helps you, was fun solving them, seemed quite original compared to other M1 questions i've done. Ask if you want anything clarified :smile:
I'm sure there's a more elegant solution to these, but I hope my method shows it's logic more fully.
(edited 12 years ago)

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