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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..
Time of flight=1.60s
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).
are u sure about this answer
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
yar solution upload kr do.
mere answer ye hain
c true bcz acording to law of conversion of momentum of an isolated sytem before and of collision is constant.
earth is inertial frame of refrence
yarrr help ker doo na pleaseee.....
about q 3 & 4....Please ...thanksss