Question

A stone 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 = 10 m/s² )​

Answer 1

Solution:

Given:

Speed of stone = 40 m/s

Distance = 40 m

Now,

We need to calculate the time interval.

Using equation of mass:

$\implies$ $\boxed{\sf{s=ut-\frac{1}{2}gt^{2}}}$

Where,

$\implies$ s = distance

$\implies$ u = speed

$\implies$ t = time

$\implies$ g = acceleration due to gravity

Substituting the values into the formula,

We get:

$\implies$ 40 = 40t - 1/2 × (-10) × t^2

$\implies$ 40 = 40t - 5t^2

$\implies$ 5t^2 - 40t + 40 = 0

$\implies$ t^2 - 8t + 8 = 0

By quadratic formula:

Note: Check this attachment!

Hence,

The time interval is: $\sf{4 \pm 2 \sqrt 2}$

____________________

Answered by: Niki Swar, Goa❤️

Answer 2

Solution :

u =40 m/s

S = 40 m

a = -g

time ,t = ?

By Newton's equation of motion ;

$\sf\:S=ut+\dfrac{1}{2}\times\:gt^{2}$

$\sf\:40=40t-\dfrac{1}{2}\times 10\times\:t^{2}$

$\sf\:40=40t-5t^{2}$

$\sf\:5t^{2}-40t+40=0$

$\sf\:t^{2}-8t+8=0$

After solving you get

$\sf\:t=\dfrac{8+\sqrt{32}}{2}$

$\sf\:t=4+2\sqrt{2}$

$\sf\:t=2.8+4$

$\bf\:t=6.8\:sec$

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\dfrac{a}{2}(2n-1)$