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
Gió
Jan 6, 2016

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#

Max
Jan 6, 2016

#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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