# An astronaut on Jupiter drops a rock straight downward from height of 1.20m. If the acceleration of gravity on Jupiter is 24.8m/s^2 what...?

is the speed of the rock just before it lands?

### 3 Answers

- billrussell42Lv 71 month agoFavorite Answer
v = √(2gh)

v = √(2•24.8•1.2) = 7.71 m/s

falling object starting from rest

h is height in meters, t is time falling in seconds,

g is acceleration of gravity, usually 9.8 m/s²

v is velocity in m/s

h = ½gt²

t = √(2h/g)

v = √(2gh)

h = v²/2g

v = gt

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- MyRankLv 61 month ago
Given,

Height (h) = 1.20m

Acceleration due to gravity (g) = 24.8m/sec²

Initial speed (u) = 0m/sec [∵ freely falling]

Final speed (v) = ?

Using kinematic relation:-

v² = u² + 2gh

= v² = (0m/sec)² + 2 x 24.8m/sec² x 1.20m

= v = √2 x 24.8 x 1.20 = 7.71m/sec

∴ Final speed (v) = 7.71m/sec.

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- SpacemanLv 71 month ago
a = gravitational acceleration on Jupiter = 24.8 m/s²

d = distance of the rock's fall = 1.20 m

v = final velocity of the rock = to be determined

v = √(2ad)

v = √(2·24.8 m/s²·1.20 m)

v = 7.714920609 m/s ≈ 7.71 m/s

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