Belajar Bareng Soal Fisika Saintek SBMPTN #1

h = 1/2 GT^2

h = 1/2 . 10 . 4

h = 20 m

s = 80+0+1/2(-10)(2)^2

   =80-20

  =60

S= 80 + V₀+ ½ (-10).(2)² 

= 80-20= 60

80 - (0+1/2(10)2²) = 80 - 20 = 60

h_t = h₀ - ½ gt²

     ^{= 80 - ½ 10.4}

    = 80 - 20

    = 60

h1= 80 m

t=2 s

g=10 m/s

v0= 0 m/s karena GJB

h2....?

jawab

s= v0.t+1/2 gt^2

s= 0+ 1/2 . 10. 2.2

s= 20 m

h=80-20=60

c

h=1/2 GT^2

h=1/2.10/2

h=10m

60 m

h/t=80/2=40

E. 60M

80-20 = 60

Vt = Vo + g.t = 0 + 10.2 = 20m/s

Vt^2 = Vo^2 - 2.g.h

400 = 0 - 2.10.h

400 = -20h

h = -20m

80 - 20 = 60m

40

h2-h1=1/2gt²

80-h1=1/2.10.2²

h1=80-20 m

h1= 60 m

C.40 M

Vt = gt = 20 

Em di titik awal = mgh = m.10.80 = 800m

Em di titik A (setelah 2s) :

Em = Ek + Ep

Ep = Em - Ek

Ep = 800 m - 1/2 m.20.20

Ep = 800m - 200m

mgh = 600m

10 mh = 600m

h = 60 m

E. 60 m

E.60

h1 = vo + 1/2 at ^2

 karena gerak jatuh bebas maka vo=0

h1= 1/2 at ^2

    = 1/2 10 .4

    = 20 meter

h = h0 - h1 

   = 80 - 20 

   = 60 meter

20

h = 1/2gt²

h =1/2.10.2²

h = 1/2.40

h = 20 m =  dh = 80 - 20 = 60

h=1/2.g.t^2                   h'=H-h

h=1/2.10.2^2                h'=80-20

h=20 m                         h'=60 m