Respuesta :
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 + [tex]\frac{1}{2}[/tex] 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) + [tex]\frac{1}{2}[/tex] g(0.63)²
0.72 = [tex]\frac{1}{2}[/tex] g(0.63)²
g = [tex]\frac{2\times0.72}{(0.63)^2}[/tex]
= 3.63 meter/sec²
Answer:
3.63 meter/sec²
Step-by-step explanation:
If a rock fall from a height h, then
[tex]h=ut+\frac{1}{2}gt^2[/tex]
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.
[tex]0.72 = (0) \times (0.63) + \frac{1}{2}g(0.63)^2[/tex]
[tex]0.72 = \frac{1}{2}g(0.63)^2[/tex]
Multiply both sides by 2.
[tex]1.44= 0.3969g[/tex]
Divide both sides by 0.3969.
[tex]\frac{1.44}{0.3969}=g[/tex]
[tex]g\approx 3.63[/tex]
Therefore, the magnitude of the acceleration due to gravity on the planet is 3.63 meter/sec².
