Who's dropping these balls?

Classical Mechanics Level pending

A ball at rest is dropped from the top of a building. It takes $0.2 \text{ s}$ to pass a window of height $4.2 \text{ m}$ . How far is the top of the window from the top of the building?

Take $g = 9.81 \text{ m/s}^2$ .

The answer is 20.

1 solution

Lu Chee Ket
Oct 25, 2015

v is increasing constantly with constant gravitational pull regardless of air friction. Hence, speed at the middle of window is the average distance to be taken;

Average speed = speed at 2.1 m down from top of window = 4.2 m/ 0.2 s = 21 m/ s;

v = u + a t => 21 = 0 + 9.81 t => t = 21/ 9.81;

x = u t + (1/ 2) a t^2 => x = 0 + (1/ 2) 9.81 (21/ 9.81)^2 = 22.477064220183486238532110091743;

x - 2. 1 = 20.377064220183486238532110091743.

With Int (x - 2.1) = 20, t = 2.1226414025819688152912344426131 instead,

v' = 20.823112159329114078007009882035

h' = 4.1646224318658228156014019764069 which is all right.

Answer: 20 m

Who's dropping these balls?

The answer is 20.

1 solution

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