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# word problem

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you and your brother are playing air hockey on the kitchen table, using your hands and your mother's china sugar bowl lid as the puck. you smack is towards your brother and he misses the block. the china lid slides off the table going 5m/s and sails across the kitchen. if the table is 1.2m high, how long will it take before the lid shatter on the ground?

Jul 3, 2014

#2
+28029
+13

The 1.2m refers to the height above the ground, and the time it takes to shatter is just the time taken to fall to the ground. Since the puck starts with zero vertical velocity this can be obtained from the following (if we ignore air resistance):

h = (1/2)gt2, where h = 1.2m, g = 9.81m/s2, and t is the time.

Rearranging, we have t = √(2h/g)

$${\mathtt{t}} = {\sqrt{{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{1.2}}}{{\mathtt{9.81}}}}}} \Rightarrow {\mathtt{t}} = {\mathtt{0.494\: \!619\: \!366\: \!829\: \!497\: \!9}}$$

t ≈ 0.495 seconds.

The horizontal velocity is irrelevant in determining the time.  It becomes relevant if we want to know how far away from the table it got before it shattered!

Jul 3, 2014

#1
+3

First off let's see how far the lid has to travel before it falls off the kitchen table.  Well the table is 1.2m long, so the distance the lid has to travel would be 1.2m.

Okay, how long does it take the lid to travel that distance?  Let's say you're walking at a pace of 5m/s.  How long does it take you to walk 10m?  2 seconds!  What we just did was divide the distance that we needed to travel by our speed, so let's do that with the lid.

Distance = 1.2m

Speed = 5m/s

We know that time=distance/speed ==> time=(1.2m/(5m/s)).

Jul 3, 2014
#2
+28029
+13

The 1.2m refers to the height above the ground, and the time it takes to shatter is just the time taken to fall to the ground. Since the puck starts with zero vertical velocity this can be obtained from the following (if we ignore air resistance):

h = (1/2)gt2, where h = 1.2m, g = 9.81m/s2, and t is the time.

Rearranging, we have t = √(2h/g)

$${\mathtt{t}} = {\sqrt{{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{1.2}}}{{\mathtt{9.81}}}}}} \Rightarrow {\mathtt{t}} = {\mathtt{0.494\: \!619\: \!366\: \!829\: \!497\: \!9}}$$

t ≈ 0.495 seconds.

The horizontal velocity is irrelevant in determining the time.  It becomes relevant if we want to know how far away from the table it got before it shattered!

Alan Jul 3, 2014
#3
+101769
+10

I am going to do it using calculus.

$$\\ When\:\: t=0,\:\: y=1.2m,\:\: \dot{y}=0 \\\\ \ddot{y}=-9.8m/s^2 \:\:(gravity)\\\\ \dot{y}=-9.8t+c_1\qquad \mbox{But since y dot=0 when t=0,} c_1=0}\\\\ \dot{y}=-9.8t\\\\ y=\frac{-9.8t^2}{2}+c_2 \qquad \mbox{substituting you get }c_2=1.2}\\\\ y=\frac{-9.8t^2}{2}+1.2\\\\ \mbox{Find t when x=0}\\\\$$

$$\begin{array}{rll} 0&=&\frac{-9.8t^2}{2}+1.2\\\\ 0&=&-9.8t^2+2.4\\\\ 9.8t^2&=&2.4\\\\ t^2&=&\frac{2.4}{9.8}\\\\ t&=&\sqrt{\frac{2.4}{9.8}}\\\\ t&=&0.4948\\\\ \end{array}$$

$$\mbox{The lid will shatter in 0.495 seconds}$$

.
Jul 5, 2014
#4
+101769
+10

I would like to thank the anonymous answerer for participating.  The reason that yours is not correct is that you have assumed a constant velocity of 5m/s.

The velocity is not constant.  In the beginning the downward velocity is 0.

Then as the lid falls it gets faster because it is accelerating at approx 9.8m/s2 (That is gravity)

Jul 5, 2014