Question

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 ?

Answer 1

Answer:

3.63 meter/sec²

Step-by-step explanation:

If a rock fall from a height h, then

$h=ut+\frac{1}{2}gt^2$

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.

$0.72 = (0) \times (0.63) + \frac{1}{2}g(0.63)^2$

$0.72 = \frac{1}{2}g(0.63)^2$

Multiply both sides by 2.

$1.44= 0.3969g$

Divide both sides by 0.3969.

$\frac{1.44}{0.3969}=g$

$g\approx 3.63$

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

Answer 2

Answer:

3.63 meter/sec²

Step-by-step 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 + $\frac{1}{2}$ 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) + $\frac{1}{2}$ g(0.63)²

0.72 = $\frac{1}{2}$ g(0.63)²

g = $\frac{2\times0.72}{(0.63)^2}$

= 3.63 meter/sec²