F.4 Physics Mechanics
Please help me with this question, and show steps. Thank you!
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.)
- 7 年前最佳解答
Vx stands for ball X 's velocity
Vy stands for ball Y
Vxy stands for "velocity of Y relative to X"
when the Y is released
X has displacement=1/2 * 10 * (2^2)=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"
y needs to fly for t=(2*100/g)^1/2 "-2" (released 2 s after ball X)
so the answer is 20+(2√5-2)*20=40√5-20
- LILv 77 年前
Sx = 1/2*gt^2
Sy = 1/2*g(t-2)^2
D = Sx – Sy = 1/2*gt^2 - 1/2*g(t-2)^2 = 2g(t-1)
100 = 1/2*gt^2
t = √(200/g)
D = 2g(√(200/g)-1) = 2( 10√(2g) – g )