Question

What is the time taken for a mango to fall 20m from it tree (g=10m/s)​

Answer 1

Hi there!

We can use the kinematic equation:

$d = v_0t + \frac{1}{2}at^2$

d = displacement (20 m)

v0 = initial velocity (dropped from rest, so 0 m/s)

t = time (s)

a = acceleration due to gravity (10 m/s²)

Rearrange the equation to solve for time:

$d = 0 + \frac{1}{2}at^2\\\\2d = at^2\\\\t^2 = \frac{2d}{a}\\\\t = \sqrt{\frac{2d}{a}}$

Solve using the given values:

$t = \sqrt{\frac{2(20)}{10}} = \sqrt{4} = \boxed{2 sec}$

Answer 2

Answer:

$\boxed {\boxed {\sf 2 \ seconds}}$

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:

$d= v_i t + \frac{1}{2} at^2$

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.

• $v_i$= 0 m/s
• d= 20 m
• a= 10 m/s²

Substitute the values into the equation.

$20 \ m = (0 \ m/s)(t) + \frac{1}{2} (10 \ m/s^2)(t^2)$

$20 \ m = \frac{1}{2} (10 \ m/s^2)(t^2)$

$20 \ m = (5 \ m/s^2)(t^2)$

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.

$\frac {20 \ m}{5 \ m/s^2}= \frac{(5 \ m/s^2)(t^2)}{5 \ m/s^2}$

$\frac {20 \ m}{5 \ m/s^2}=t^2$

$4 \ s^2=t^2$

The variable t is being squared. Take the square root of both sides.

$\sqrt { 4 \ s^2 }= \sqrt{t^2}$

$2 \ s=t$

It takes 2 seconds for a mango to fall 20 meters.