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# PHY101 Assignment#1 Fall 2013 Opening Date 21-11-2013

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### Replies to This Discussion

plzzzzzzzzzzz assignment ka solution bata dy koi ........ bohat confusing questions hain ..plzzzzzzzzzzzzzzzzz

plxzxxxxxxxxxxxxxxx helpppppppppppppp meeeeeeeeeeeeeeeeeeeeee! lagta ha yeh subjects koi ne parta

Handouts se read krain ye numericals ki trah hi hai is ko solve krna hai...

Formula put krain or ans nikalen..

Question: 01

Time of flight=1.60s

Range=75cm

Time of flight = 2·V·sinΘ/g = 1.60 s ← #1
range = (V²sin(2Θ))/g = 0.75 m ← #2
Divide #2 by #1:
0.75m / 1.6s = Vsin(2Θ) / 2sinΘ
But sin(2Θ)= 2sinΘcosΘ, so
0.47 m/s = VcosΘ ← initial horizontal component -- #3
rearranges to V = 0.47m/s / cosΘ

Now plug that into #1:
2 * (0.47m/s / cosΘ) * sinΘ / 9.8m/s² = 1.6 s
0.096s * tanΘ = 1.6s
tanΘ = 16.7

Now multiply both sides of #3 by tanΘ = 16.7
0.47m/s * 16.7 = VcosΘ * tanΘ = VsinΘ
VsinΘ = 7.8 m/s ← vertical component

Note that we don't need to know either the angle or the initial velocity; we just need the vertical component (VsinΘ) and the horizontal component (VcosΘ).

my answers are vi = 0.03

vy= 0.67 and i think i m rite because initial velocity components in both direction will be zero.

plz upload the correct solution of all questions

if some body knows

de 2 d ae :/ ab 2 dramay na lgao :/

vi=0.46 m/s and vf=1.38 m/s

Ok Assignment Done!

:)

Anybuddy want help feel free to ask.

+ saif ahannn gud

q.no. 1
vx=.468 ms
q no.2
a=0

q.no.3
a false
b false
c true bcz acording to law of conversion of momentum of an isolated sytem before and of collision is constant.

q.no. 4
earth is inertial frame of refrence

yarrr help ker doo na pleaseee.....

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