Certainly! Let's work through this step-by-step:
### Part (a):
To find the time it takes for the stone to fall 4.9 meters, we use the kinematic equation for objects in free fall:
[tex]\[ h = \frac{1}{2} g t^2 \][/tex]
where:
- [tex]\( h \)[/tex] is the height (4.9 meters),
- [tex]\( g \)[/tex] is the acceleration due to gravity (9.8 m/s²),
- [tex]\( t \)[/tex] is the time in seconds.
Rearranging to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \sqrt{\frac{2h}{g}} \][/tex]
Substitute the known values:
[tex]\[ t = \sqrt{\frac{2 \cdot 4.9 \, \text{m}}{9.8 \, \text{m/s}^2}} \][/tex]
[tex]\[ t = \sqrt{\frac{9.8 \, \text{m}}{9.8 \, \text{m/s}^2}} \][/tex]
[tex]\[ t = \sqrt{1} \][/tex]
[tex]\[ t = 1 \, \text{s} \][/tex]
So, it takes [tex]\( 1 \)[/tex] second for the stone to fall 4.9 meters.
### Part (b):
Next, we want to find the velocity of the stone at the end of the fall. The velocity of an object falling from rest under gravity can be found using:
[tex]\[ v = g t \][/tex]
where:
- [tex]\( v \)[/tex] is the velocity,
- [tex]\( g \)[/tex] is the acceleration due to gravity (9.8 m/s²),
- [tex]\( t \)[/tex] is the time in seconds (from part a, [tex]\( t = 1 \)[/tex] second).
Substitute the known values:
[tex]\[ v = 9.8 \, \text{m/s}^2 \cdot 1 \, \text{s} \][/tex]
[tex]\[ v = 9.8 \, \text{m/s} \][/tex]
So, the stone moves at [tex]\( 9.8 \, \text{m/s} \)[/tex] at the end of the fall.
### Part (c):
Finally, to determine the acceleration of the stone after 3 seconds, we recognize that the only force acting on the stone is gravity (assuming no air resistance). The acceleration due to gravity is constant.
Thus, the acceleration [tex]\( a \)[/tex] after any amount of time, including 3 seconds, remains:
[tex]\[ a = 9.8 \, \text{m/s}^2 \][/tex]
So, the acceleration of the stone after 3 seconds is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
### Summary:
(a) The time to fall 4.9 meters: [tex]\( 1 \)[/tex] second
(b) The velocity at the end of that fall: [tex]\( 9.8 \, \text{m/s} \)[/tex]
(c) The acceleration after 3 seconds: [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]
