Author Topic: Kinematics (Read 1266 times) Tweet Share

nerd · « **on:** November 29, 2008, 09:42:41 pm »

Please help me with the attached question...

shinny · « **Reply #1 on:** November 29, 2008, 10:40:52 pm »

This one's quite tricky..
Intuitively, define O to be s=0, the vertically upwards to be positive s, and vertically downwards to be negative s. Also, let $t_1$ be when the stone hits the lift, and let $s_1$ be the depth at which the stone hits the lift. Therefore we have for the stone; $u=14$ , $a=-9.8$ , $t=t_1-5$ , $s=s_1$ , while for the lift, we have $u=-3.5$ , $a=0$ , $t=t_1$ , $s=s_1$

Using $s=ut+\frac{1}{2}at^{2}$ , we get $14(t_1 -5)+\frac{1}{2} (-9.8) (t_1 -5)^{2}=-3.5t_1$
Solving on a CAS or solver or whatever; we get $t_1=4.19, 9.39$

But remember how we defined t of the stone as $t_1-5$ ? This means we reject 4.19 as the answer as this would lead to a negative time, and take 9.39

Subbing $t_1=9.39$ , we get $s=(-3.5)(9.39)=-32.85$
Therefore the answer is 33 metres (I hope; kinematics has never been my strong point)

nerd · « **Reply #2 on:** November 29, 2008, 10:50:08 pm »

Wow! That's right shinjitsuzx! Thanks so much.

That sure is a complicated question....

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Author Topic: Kinematics (Read 1266 times) Tweet Share

nerd

Kinematics

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