Physics starship

Anonymous

2012-10-02 7:46 pm

Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance D from the foot of the table. This starship now lands on the
unexplored Planet X. The commander, Captain Curious, rolls the same ball off the
same table with the same initial speed as on earth and finds that it lands a
distance 2.82 D from the foot of the table.

What is the acceleration due to gravity on Planet X?

gX = ? m/s^2

回答 (1)

天同

2012-10-02 8:39 pm

✔ 最佳答案

Let h be the height of the table above the floor, and v be the projecting speed of the ball.

On earth:
Consider the horizontal motion: D = vt where t is the time of flight
i.e. t = D/v
Vertical motion: use s = ut + (1/2)at^2
with s = h, u = 0 m/s, a = g
h = (1/2)gt^2
i.e. h = (g/2).(D/v)^2 ---------------- (1)

On Planet X,
Horizontal motion: 2.82D = vt' where t' is the time of flight on the planet
t' = 2.82/v
Vertical motion: h = (1/2)g't'^2 where g' is the acceleration due to gravity on the planet
hence, h = (g'/2).(2.82/v)^2 -------------- (2)

Equating (1) and (2),
(g/2).(D/v)^2 = (g'/2).(2.82D/v)^2
g' = g/2.82^2 = 1.26 m/s^2 (taking g = 10 m/s^2)

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