# F.4 Physics Mechanics

Two identical balls X and Y are released from rest from the top of a building 100 m above the ground. Ball Y is released 2 s after ball X. What is the maximum vertical separation between balls X and Y in the air? (Neglect air resistance.)

### 2 個解答

• 最佳解答

Vx stands for ball X 's velocity

Vy stands for ball Y

Vxy stands for "velocity of Y relative to X"

Vxy=Vy-Vx

Acceleration=g=10 m/s^2

when the Y is released

X has displacement=1/2 * 10 * (2^2)=20

Vxy=Vy-Vx=Vy-0=Vy=10*2=20

Axy=Ax-Ay=g-g=0 (acceleration of xy)

as a result Vxy is const =20

so maximum vertical separation between balls X and Y in the air is

"when the X touches ground"

it take

t=(2*100/g)^1/2

y needs to fly for t=(2*100/g)^1/2 "-2" (released 2 s after ball X)

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• 從位移看:

Sx = 1/2*gt^2

Sy = 1/2*g(t-2)^2

兩者相距

D = Sx – Sy = 1/2*gt^2 - 1/2*g(t-2)^2 = 2g(t-1)

所以t越大D越大

直到X撞地

100 = 1/2*gt^2

t = √(200/g)

所以

D = 2g(√(200/g)-1) = 2( 10√(2g) – g )

若g=10

D= 40√5-20

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