Editorial - Fisika - Gerak Parabola

The General Formulae

(v_x)=(v_x(0))=(v_0)×cos(θ)

(v_y)=(v_0)*sin(θ)-g*t

(y_max)=((v_0)2*(sin^2)(θ))/2g

(t_(y_max))=((v_0)*sin(θ))/g

(x_max)=((v_0)2*sin(2)*θ)/g=((v_0)2×2*sin(θ)*cos(θ))/g

(t_(x_max))=(2*(v_0)*sin(θ))/g=2×(t_(y_max))

and other variable motion equations.

Kemampuan Awal

Problem 1

(x_max)=(102⋅2*sin(37)*cos(37))/10=20⋅3/5⋅4/5=24/25⋅10=9.6

Problem 2

When it's at the peak, the only component is the horizontal component.

(v_x)=25⋅2/5=10

Problem 3

x=10⋅cos(30)=10⋅√(,3)/2=5√(,3)

y=10⋅sin(30)-10/2=10⋅1/2-5=0

∴(5√(,3),0)

Problem 4

2000=(10*t2)/2

t2=4000/10=400

t=20

(s_x)=20⋅250=5000=5

Problem 5

12=(v_0)⋅4/5⋅t

t=15/(v_0)

4=(v_0)⋅3/5⋅t-5*t2

4=(v_0)⋅3/5⋅15/(v_0)-5⋅225/((v_0)2)

4=9-(5⋅225)/((v_0)2)

(v_0)2=225

(v_0)={-15,+15}

EZ

Problem 6

4=t⋅(v_0)

0.2=5*t2

t2=0.2/5=1/25

t=1/5

(v_0)=4⋅5=20=7*2

Problem 7

24=((v_0)2⋅2⋅3/5⋅4/5)/10

(v_0)2=24⋅25/24⋅10=250

(v_0)=5√(,10)

(y_max)=(250⋅9/25)/20=4.5

Problem 8

1:1

karena sin(120)=sin(60)

Problem 9

80=m⋅((v_0)2)/2

(v_0)2=160/m

(v_0)=(4√(,10))/√(,m)

(v_x)=(4√(,10))/√(,m)⋅1/2=(2√(,10))/√(,m)

Find the horizontal kinetic energy (it is constant)

(E_k)=20

At the peak, the vertical kinetic energy is 0

(E_p)=80-20=60

It's likely that this solution is quite convoluted and so it may be easier to just find the height at the peak

or just use ratios idk

Problem 10

0=65+5⋅(v_0)-5⋅52

(v_0)=1*2

(v_y)=10⋅5-12=38

YAY

Soal TKA

Problem 1

Height difference is 8.5-1.5=7

7=20⋅3/5⋅t-5*t2

7=12*t-5*t2

5*t2-12*t+7=0

(5*t-7)*(t-1)

t={1,+7/5}

So we know when the rock reaches the ball's altitude

We need to find whether it reaches the ball during the first or the second time

(x_1)=20⋅4/5⋅1=16

So, the rock reaches the ball from its underside because it reaches 16 before it starts going down. This means 1is false and 2 is true.

(t_(x_max))=2⋅20⋅4/5⋅1/10=2.4

So the answer is

FALSE, TRUE, FALSE

Problem 2

4.5=((v_0)2⋅9/25)/(2*g)

1=((v_0)⋅3/5)/g

g=(v_0)⋅3/5

Substitute that

4.5=((v_0)2⋅3/5⋅3/5)/((v_0)⋅3/5⋅2)=3/10*(v_0)

(v_0)=10/3⋅9/2=15

The first statement is true.

g=15⋅3/5=9

The second statement is false.

(x_max)=(225⋅2⋅3/5⋅4/5)/9=24

The third statement is true.

Problem 3

(v_0)2⋅(2*sin(60)*cos(60))/g∶(v_0)2⋅(2*sin(45)*cos(45))/g

√(,3)/2⋅1/2:√(,2)/2:√(,2)/2

√(,3):2

The first statement is false.

(sin^2)(60)∶(sin^2)(45)

3/4∶2/4

3:2

The second statement is true because the difference is 1/3 to the original

From this ratio we can also tell that the third statement is also false because they both use the sine function as the ratio

Problem 4

200

2000=(10*t2)/2

t=20

The first statement is true.

(x_max)=20⋅200=4000

The third statement is false because the tank moves away.

v=√(,2002+2002)=200√(,2)

The second statement is false.

Problem 5

2.4=(2*(v_0)⋅3/5)/10

(v_0)=12⋅5/3=20

(v_x)=20⋅4/5=16

(v_1.2)=√(,162+…)

The first statement is clearly false.

(x_max)=2.4⋅16=38.4

The second statement is false.

The third statement is true because velocity at the start is equal to velocity at the end.

Problem 6

24=20⋅cos(53)⋅t

t=24/20⋅5/3=2

The bullet reaches the point A after 2 seconds.

We know that the distance the ball needs to travel is 12

y=16⋅2-5⋅4=12

So, the first statement is true.

v=16-2⋅10=-4

The second statement is false.

(v_y)=20⋅4/5-10⋅2=-4

v=√(,42+(20⋅3/5)2)=√(,16+144)=4√(,10)

The third statement is false.