Answer:
![\boxed {\boxed {\sf 2 \ seconds}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%202%20%5C%20seconds%7D%7D)
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:
![d= v_i t + \frac{1}{2} at^2](https://tex.z-dn.net/?f=d%3D%20v_i%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
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.
= 0 m/s - d= 20 m
- a= 10 m/s²
Substitute the values into the equation.
![20 \ m = (0 \ m/s)(t) + \frac{1}{2} (10 \ m/s^2)(t^2)](https://tex.z-dn.net/?f=20%20%5C%20m%20%3D%20%280%20%5C%20m%2Fs%29%28t%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2810%20%5C%20m%2Fs%5E2%29%28t%5E2%29)
![20 \ m = \frac{1}{2} (10 \ m/s^2)(t^2)](https://tex.z-dn.net/?f=20%20%5C%20m%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2810%20%5C%20m%2Fs%5E2%29%28t%5E2%29)
![20 \ m = (5 \ m/s^2)(t^2)](https://tex.z-dn.net/?f=20%20%5C%20m%20%3D%20%285%20%5C%20m%2Fs%5E2%29%28t%5E2%29)
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.
![\frac {20 \ m}{5 \ m/s^2}= \frac{(5 \ m/s^2)(t^2)}{5 \ m/s^2}](https://tex.z-dn.net/?f=%5Cfrac%20%7B20%20%5C%20m%7D%7B5%20%5C%20m%2Fs%5E2%7D%3D%20%5Cfrac%7B%285%20%5C%20m%2Fs%5E2%29%28t%5E2%29%7D%7B5%20%5C%20m%2Fs%5E2%7D)
![\frac {20 \ m}{5 \ m/s^2}=t^2](https://tex.z-dn.net/?f=%5Cfrac%20%7B20%20%5C%20m%7D%7B5%20%5C%20m%2Fs%5E2%7D%3Dt%5E2)
![4 \ s^2=t^2](https://tex.z-dn.net/?f=4%20%5C%20s%5E2%3Dt%5E2)
The variable t is being squared. Take the square root of both sides.
![\sqrt { 4 \ s^2 }= \sqrt{t^2}](https://tex.z-dn.net/?f=%5Csqrt%20%7B%20%204%20%5C%20s%5E2%20%7D%3D%20%5Csqrt%7Bt%5E2%7D)
![2 \ s=t](https://tex.z-dn.net/?f=2%20%5C%20s%3Dt)
It takes 2 seconds for a mango to fall 20 meters.