Question

hi friends plz help me ​

Answer 1

Answer :-

So the body is thrown vertically upward and it reaches it's maximum height in 5 sec.

Then by

v = u + at

We will find out the initial velocity

As we know that at maximum height v = 0

$\implies 0 = u + (-10)(5)$

$\implies u = 50 \: m/s$

Now we know that distance traveled in nth sec

$= u + \dfrac{1}{2}.a.(2n - 1)$

Now by checking option (1)

For 1st sec :-

$S_1 = 50 + \dfrac{1}{2}(-10).(2(1) - 1)$

$S_1 = 50 + (-5).(1)$

$S_1 = 50 - 5$

$S_1 = 45$

For 10th sec

$-S_{10} = 50 + \dfrac{1}{2}(-10).(2(10) - 1)$

$-S_{10} = 50 + (-5).(19)$

$-S_{10} = 50 - 95$

$S_{10} = 95 - 50$

$S_{10} = 45$

So distance traveled in 1st sec = Distance traveled in 10th sec.

Hence option 1 is correct.