free falling body

8/6/2019 Free Falling Body

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FREE FALLING BODY


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FREE FALLING BODIES

FREE- FALL describes any motion of abody where gravity is the only ordominant force acting upon it.

g = 9.81m/s = 32ft/s


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INITIAL VELOCITY IS EQUAL TO ZERO

V=V+gt

V=gt

S= Vtgt

S=H=gt

V=V2as

V=2gh


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PROBLEM 1

A stone is droppedfrom rest from aheight of 400ft. a.)

how long will it takefor the stone toreach the ground?b.) what is the

velocity that instantit this the ground?

400m


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Solution:

a) H=gt400ft =(32ft/s)(t) t=25 t=5 sec

b) V=V2ah

V=(0m/s)2(32ft/s)(400ft) V=25600 V=160ft/s


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PROBLEM 2

An object is thrownvertically upwardfrom the ground

with a velocity of49m/s. a) what willbe its maximumheight? b) after how

many seconds will itreturn the ground?


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Solution:

V=V+gt

0m/s=49m/s+(9.81m/s )t

t=5 sec

a) H= Vtgt H= 49m/s(5s)-

(9.81m/s)(5s)

H=122.38m

b) H= Vgt

122.38m= (9.81m/s )t

t=5 sec

Total Time = Time(up) +Time(down)

= 5sec +5sec

Total Time= 10 seconds


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PROBLEM 3

A stone is thrownvertically into the airfrom a tower 100ft.At the same instantthat a second stoneis thrown upwardfrom the ground. Theinitial velocity of the

first stone is 50ft/sand the second stoneis 75ft/s. When andwhere will the stonesbe at the same

height from the


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Solution:

Stone 1H= Vtgt =50ft/s(t)-

(32ft/s)(t)H=50t-16t

Stone 2H= Vtgt = 75ft/s(t)-

(32ft/s)(t)

H=75t-16t

Subtract from H=50t-16t-H=75t-16t H- H=-25t H- H=25t

-


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From the figureH- H=100ft

From substitute H- H=25t100ft=25t

25 25 t= 4sec

Substitute t in and

H=50t-16t

H=50(4)-16(4)H=-56ft

H=75t-16tH=75(4)-16(4)H=44ft


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PROBLEM 4

A ball is dropped from thetop of a cliff 150m high atthe same instant anotherball is thrown upwardfrom the ground. The

balls pass each other at apoint 60m along theground. What was theinitial velocity of thesecond ball?

150m


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Solution:

V=0m/s V=0m/s

H=90m

H=60mg=9.81m/s

g=9.81m/st=? t=?

t= tH= Vtgtt=H(2)/gt=90m(2)/9.81m/s

H= Vtgt60m=V(4.28s)-

(9.81m/s)(4.28)

V=35.01m/s


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PROBLEM 5

During a test a rocket istravelling upward at75m/s, and when it is40m from the ground

its engine fails.Determine themaximum height Hreached by the rocketand its speed just

before it hits theground. While inmotion the rocket issubjected to a constantdownward acceleration

of 9.81m/s due toravit .


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Solution:

V=V-2gH(Om/s)=(75m/s)-2(9.81m/s)HH=286.70m

H+40m=HH=286.70m+40mH=326.70m

V=2gHV=2(9.81m/s)(326.70m)V=80.06m/s

free falling body

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