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A rock falls from rest a vertical distance of 0.72 meters to the surface of a planet in 0.63 seconds. The magnitude of the acceleration due to gravity on the planet ?

User Maqueda
by
7.4k points

2 Answers

1 vote

Answer:

3.63 meter/sec²

Explanation:

If a rock fall from a height h, then


h=ut+(1)/(2)gt^2

Where g is the acceleration due to gravity, u is initial velocity and t is time in seconds in which the rock reaches the surface.

It is given that a rock falls from rest a vertical distance of 0.72 meters to the surface of a planet in 0.63 seconds.

h = 0.72 meters

u = 0

t = 0.63 seconds

Substitute the given values in the above formula to find the value of g.


0.72 = (0) * (0.63) +  (1)/(2)g(0.63)^2


0.72 = (1)/(2)g(0.63)^2

Multiply both sides by 2.


1.44= 0.3969g

Divide both sides by 0.3969.


(1.44)/(0.3969)=g


g\approx 3.63

Therefore, the magnitude of the acceleration due to gravity on the planet is 3.63 meter/sec².

User Falmarri
by
8.1k points
5 votes

Answer:

3.63 meter/sec²

Explanation:

When a rock fall from a height h with initial velocity u and the rock reaches the surface in t seconds the expression that represents this is

h = 4t +
(1)/(2) gt²

Where g is the acceleration due to gravity

Here h = 0.72 meters

u = 0

t = 0.63 seconds

from the given formula

0.72 = 0 ×(0.63) +
(1)/(2) g(0.63)²

0.72 =
(1)/(2) g(0.63)²

g =
(2*0.72)/((0.63)^2)

= 3.63 meter/sec²

User Finola
by
7.8k points

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