Question

A stone is thrown vertically upward with a speed of 40 m/s. The time interval for which particle was above 40 m from the ground is(g=10m/s²)

Answer 1

Explanation:

s = ut - 1/2 gt²

40 = 40t - 1/2 × 10 × t²

40 = 40t - 5t²

5t² - 40t + 40 = 0

t² - 8t + 8 = 0

by shridharacharya method

t = 4 + root8

t = 4 + 2 × 1.414

t = 4 + 2.828

t = 6.828 sec

Answer 2

Answer:

The time interval is [1.071 s, 14.92 s].

Explanation:

Given that,

Speed of stone = 40 m/s

Distance = 40 m

We need to calculate the time interval

Using equation of mass

$s=ut-\dfrac{1}{2}gt^2$

Where, s = distance

u = speed

t = time

g = acceleration due to gravity

Put the value into the formula

$40=40t-\dfrac{1}{2}\times10\times t^2$

$5t^2-80t+80=0$

$t^2-16t+16=0$

The time interval is

$[t_{1},t_{2}]=[1.071\ s,14.92\ s]$

Hence, The time interval is [1.071 s, 14.92 s].