Question

A stone is thrown vertically upward with a speed of 40m/s. The time interval for which particle was above 40 m from the ground is (g=10m/s*2)

(A) 4√2s
(B) 8s
(C) 4s
(D) 2√2s

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time\:taken=4-2\sqrt{2}\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a stone is thrown vertically upward with a speed of 40m/s.

• We have to find the time interval for which particle was above 40 m from the ground.

$\underline \bold{Given : } \\ \implies Initial \: velocity(u) = 40 \: m/s \\ \\ \implies Distance(s) = 40 \: m \\ \\ \implies Acceleration(a) = - 10 \: m /{s}^{2} \\ \\ \underline \bold{To \: Find : } \\ \implies Time \: taken(t) = ?$

$\bold{By \: Third \: equation \: of \: motion : } \\ \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \bold{ putting \: given \: values : } \\ \implies 40 = 40 \times t + \frac{1}{ \cancel2} \times ( \cancel{- 10}) \times {t}^{2} \\ \\ \implies 40 = 40t - 5 {t}^{2} \\ \\ \implies {5t}^{2} - 40t + 40 = 0 \\ \\ \implies {t}^{2} - 8t + 8 = 0 \\ \\ \bold{by \: quadratic \: formula : } \\ \implies t = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \implies t = \frac{ - ( - 8) \pm \sqrt{ {( - 8})^{2} - 4 \times 1 \times 8} }{2 \times 1} \\ \\ \implies t = \frac{8 \pm \sqrt{64 - 32} }{2} \\ \\ \implies t = \frac{8 \pm 4 \sqrt{2} }{2} \\ \\ \bold{\implies t =4 \pm 2 \sqrt{2} \: sec} \\ \\ \bold{ We \: take \: minimum \: time } \\ \bold{\therefore Time \: taken \: to \: reach \: 40 \: m \: is \: 4 - 2 \sqrt{2} \: sec}$

Answer 2

Solution:

Given:

=> Initial velocity (u) = 40 m/s

=> Distance (s) = 40 m

=> Acceleration (a) = - 10 m/s²

To Find:

=> Time (t)

Formula used:

$\sf{\implies s = ut+\dfrac{1}{2}at^{2}}$

So, By using 3rd equation of motion,

$\sf{\implies s = ut+\dfrac{1}{2}at^{2}}$

$\sf{\implies 40 = 40 \times t + \dfrac{1}{2}\times (-10) \times t^{2}}$

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

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

Take 5 common from above equation, we get

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

By using Quadratic equation, because here splitting middle term not works.

$\sf{\implies x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Here x is replace by t.

$\sf{\implies t=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Where,

=> a = 1

=> b = -8

=> c = 8

So, put the values in the above formula,

$\sf{\implies t = \dfrac{-(-8)\pm \sqrt{(-8)^{2}-4\times 1\times 8}}{2}}$

$\sf{\implies t=\dfrac{8\pm \sqrt{64-32}}{2}}$

$\sf{\implies t=\dfrac{8\pm \sqrt{32}}{2}}$

$\sf{\implies t=\dfrac{8\pm 4\sqrt{2}}{2}}$

$\large{\boxed{\boxed{\red{\sf{\implies t=4\pm 2\sqrt{2}\;sec}}}}}$

Hence, Time taken = 4 ± 2√2 sec