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Assignment 1

The document presents solutions to three physics problems involving differential equations and projectile motion. It calculates the values of constants A and B for a motion equation, determines the time of flight and horizontal distance for a projectile, and discusses conditions for weightlessness in a rotating system. Key results include the angle of projection and the radius of rotation for achieving weightlessness.

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jasmeetcool1234
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11 views3 pages

Assignment 1

The document presents solutions to three physics problems involving differential equations and projectile motion. It calculates the values of constants A and B for a motion equation, determines the time of flight and horizontal distance for a projectile, and discusses conditions for weightlessness in a rotating system. Key results include the angle of projection and the radius of rotation for achieving weightlessness.

Uploaded by

jasmeetcool1234
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Using initial condition

Sol. 1 a = Kt −  2 x
at t = 0
d 2x
2
=a x = 0, x&= 0,
dt
d 2x Calculating the value of A and B
2
= Kt −  2
x
dt A=0
On solving the differential equation −K
B= 3

x = C.F . + P.I . Kt K sin(  t )
x= −
C.F . = A cos(  t ) + B sin(  t )  2
3

Kt
P.I . =
2
Kt
x = A cos(  t ) + B sin(  t ) +
2
Sol.2 v = u + at
0 = v0 cos( ) − gt
v0 cos( )
t=
g
2v0 cos( )
Total time of flight= 2t =
g y
Horizontal distance travelled by ball
1
s = ut + at 2 x
2
2
2v0 cos( ) 1  2v0 cos( ) 
s = v0 sin( )  −  0.4   
g 2  g 
s=0 ball return back to boy hand

0.2  4
tan( ) =
2 g
 = 2.33490
Sol.3 a = an eˆn + at eˆt
v2 eˆt
a= eˆn + v&eˆt

eˆn
For weightlessness 𝑎𝑛 = 𝑔

v2
9.79 =

 = 5044.1987 m
v
=
&

&= 2.5241deg/ s

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