Question

answer if u can. rrdhk ​

Answer 1

hope this will be helpful to you

Answer 2

Answer:

$\bold\red{(1)4\sqrt{2}\:s}$

Step-by-step explanation:

It is being given that,

When, the body is thrown vertically upward,

acceleration, a = -g = -5 $m{s}^{-2}$

Initial speed , u = 40 m/s

and displacement, s = 40 m

Let, time taken be 't' seconds.

So, by the kinematic formula,

$s = ut + \frac{1}{2} a {t}^{2}$

we get,

$= > 40 = 40t + \frac{1}{2} \times ( - 5) {t}^{2} \\ \\ = > 40 = 40t - 5 {t}^{2} \\ \\ = > 5 {t}^{2} - 40t + 40 = 0 \\ \\ = > {t}^{2} - 8t + 8 = 0$

Now, solving the quadratic equation for t,

we get,

$= > t = \frac{8± \sqrt{64 - 32} }{2} \\ \\ = > t = \frac{8±4 \sqrt{2}} {2} \\ \\ = > t = 4±2 \sqrt{2}$

Now, we have ,

Total time taken to reach the maximum height,

T = u/g

=> T = 40/10 = 4 seconds

Since, total time taken to reach maximum height is 4 seconds

therefore, time taken to reach the height of 40 m will be,

=> t = (4-2√2) seconds.

So, the time interval for which particle was above 40 m is equal to ,

T - t = [4-(4-2√2)] seconds = 2√2 seconds.

Now, it is the time during upward motion.

The time taken during return journey or downward motion till 40 m above ground will also be 2√2 Seconds.

Hence, the total time is (2√2+2√2) second ,

that is,

(1) 4√2 seconds