Answer:
[tex]\boxed {\boxed {\sf 2 \ seconds}}[/tex]
Explanation:
We are asked to find the time it takes for a mango to fall 20 meters.
We know the distance, acceleration, and initial velocity, so we will use the following kinematic equation:
[tex]d= v_i t + \frac{1}{2} at^2[/tex]
The mango is dropped from rest, so the initial velocity is 0 meters per second. It falls a distance of 20 meters. The acceleration due to gravity is 10 meters per second squared.
- [tex]v_i[/tex]= 0 m/s
- d= 20 m
- a= 10 m/s²
Substitute the values into the equation.
[tex]20 \ m = (0 \ m/s)(t) + \frac{1}{2} (10 \ m/s^2)(t^2)[/tex]
[tex]20 \ m = \frac{1}{2} (10 \ m/s^2)(t^2)[/tex]
[tex]20 \ m = (5 \ m/s^2)(t^2)[/tex]
We are solving for time, so we must isolate the variable t. It is being multiplied by 5 meters per second squared. The inverse operation of multiplication is division, so divide both sides by 5 m/s².
[tex]\frac {20 \ m}{5 \ m/s^2}= \frac{(5 \ m/s^2)(t^2)}{5 \ m/s^2}[/tex]
[tex]\frac {20 \ m}{5 \ m/s^2}=t^2[/tex]
[tex]4 \ s^2=t^2[/tex]
The variable t is being squared. Take the square root of both sides.
[tex]\sqrt { 4 \ s^2 }= \sqrt{t^2}[/tex]
[tex]2 \ s=t[/tex]
It takes 2 seconds for a mango to fall 20 meters.
