Answer:
[tex]\boxed {\boxed {\sf 12.0 \ seconds}}[/tex]
Explanation:
We are asked to find the time it tasked for a ball to hit the ground. We know the height, acceleration, and initial velocity, so we will use the following kinematic equation:
[tex]d= v_i t + \frac{1}{2} at^2[/tex]
The ball starts at rest, so its initial velocity is 0 meters per second. The distance is 700 meters. The ball is in free fall, so the acceleration is due to gravity, or 9.8 meters per second squared.
- [tex]v_i[/tex]= 0 m/s
- d= 700 m
- a= 9.8 m/s²
Substitute the values into the formula.
[tex]700 \ m = (0 \ m/s)*t + \frac{1}{2} (9.8 \ m/s^2)t^2[/tex]
[tex]700 \ m = \frac{1}{2} (9.8 \ m/s^2)t^2[/tex]
[tex]700 \ m = 4.9 \ m/s^2 * t ^2[/tex]
Divide both sides by 4.9 meters per second squared to isolate the variable t.
[tex]\frac {700 \ m}{4.9 \ m/s^2}= \frac{4.9 \ m/s^2*t^2}{4/9 \ m/s^2}[/tex]
[tex]\frac {700 \ m}{4.9 \ m/s^2}=t^2[/tex]
[tex]142.857142857 \ s^2=t^2[/tex]
Take the square root of both sides.
[tex]\sqrt {142.857142857 \ s^2} =\sqrt{t^2[/tex]
[tex]11.9522860933 \ s = t[/tex]
The original measurement has 3 significant figures, so our answer must have the same.
3 sig fig for our answer is the tenths place. The 5 in the tenths place tells us to round the 9 to a 0, then round the 1 to a 2.
[tex]12.0 \ s = t[/tex]
It takes the ball approximately 12.0 seconds to hit the ground.
