Question

Physics Problem Solving

a.) A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin's displacement between ?1=0.460s and t2=1.15 s?

b.) You attach a meter stick to an oak tree, such that the top of the meter stick is 2.07 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.136 seconds to pass the length of the meter stick, how high above the ground in meters was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on the way down?

Answer 1

(a) Use equation $s=ut+1/2at^{2}$ where u is the initial velocity which is Zero(initially at rest) and acceleration is g

so $s=1/2gt^{2}$

-------------------

In $t_{1}=0.46s$ coin falls a height of $s_{1}$

$s_{1}=1/2gt_{1}^{2}$

$s_{1}=0.5*(9.81m/s^{2})*(0.46s)^{2}$

$s_{1}=1.0379m$

------------------------

In $t_{2}=1.15s$ coin falls a height of $s_{2}$

$s_{2}=1/2gt_{2}^{2}$

$s_{2}=0.5*(9.81m/s^{2})*(1.15s)^{2}$

$s_{2}=6.4869m$

-----------

Required answer = $s_{2}-s_{1}=6.4869m-1.0379m$

ANSWER: = $5.449m$

===========================

(b) In time $t_{1}$ ,the acorn falls from a height $s_{1}$ and reach top of meter stick .

After 0.136s ,the acorn falls 1 meter and reach bottom of meter stick covering distance $s_{2}\ where\ (s_{2}=1m +s_{1})$ .

-------------

Find $s_{1}$

Use equation $s=ut+1/2at^{2}$ where u is the initial velocity which is Zero(initially at rest) and acceleration is g

so $s=1/2gt^{2}$

$s_{1}=1/2gt_{1}^{2}$ --------------------(1)

---------------------

$s_{2}=1/2gt_{2}^{2}$

$1+s_{1}=1/2gt_{2}^{2}$

From (1)

$1+1/2gt_{1}^{2}=1/2gt_{2}^{2}$

$1+1/2gt_{1}^{2}=1/2g(0.136+t_{1})^{2}$

$1+0.5*9.81t_{1}^{2}=0.5*9.81(0.136+t_{1})^{2}$

$(a+b)^{2}=a^{2}+2ab+b^{2}$

$1+4.905t_{1}^{2}=4.905(0.018496+0.272t_{1}+t_{1}^{2})$

$1+4.905t_{1}^{2}=0.09072288+1.33416t_{1}+4.905t_{1}^{2}$

$1=0.09072288+1.33416t_{1}$

$(1-0.09072288)/1.33416=t_{1}$

$t_{1}=0.6815s$

--------

$s_{1}=1/2gt_{1}^{2}$

$s_{1}=0.5*9.81*0.6815^{2}$

$s_{1}=2.2781m$

=========

so the acorn falls from $2.2781m$ above the meter stick

so height from ground =$2.2781m +2.07m$

ANSWER: =$4.3481m$

===================