In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the shell cracks. If a seagull flies to a height of 27 m, how long will the clam take to fall?

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In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the shell cracks. If a seagull flies to a height of 27 m, how long will the clam take to fall?

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Answer 1 · 2016-01-06T17:09:19

I found: $2.3s$

Explanation:

The shell will fall with zero initial velocity (dropped) and under the action of acceleration of gravity $g$ .
We can use:
$color(red)(y_f-y_i=v_it+1/2*at^2)$

$0-27=0t-1/2*9.8t^2$
$t=sqrt((2*27)/9.8)=2.3s$

Answer 2 · 2016-01-06T17:13:57

$2.3s$ (2sf), Assuming dropped from rest vertically with no air resistance.

Explanation:

Assuming that we are treating the clam as falling vertically from rest with no air resistance, the following should sort it:

Downwards is positive.

$s=27m$
$u=0ms^-1$
$v=vms^-1$
$a=g=9.81ms^-2$
$t=?s$

We are looking for $t$ , and with the information we have, we don't need to worry about $v$ .

Use the constant acceleration equation:
$s=ut+1/2at^2$

$27=1/2(g)t^2$

$t^2=54/9.81$

$t=2.34618....s$

$t=2.3s$ (2sf)

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In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the shell cracks. If a seagull flies to a height of 27 m, how long will the clam take to fall?

2 Answers

Explanation:

Explanation:

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