Question

a particle is thrown vertically up such that distance travelled by it in 2nd second and 9th second is same. finda the maximum height upto which the particle rises​

Answer 1

the formula for s at any second is given at the top...

then further calculation is easy

Answer 2

The height upto the particle is thrown is 125m.

Assume,

• 2nd time distance travelled as

t = 1 to t = 2

• The nth time distance travelled as

t = n - 1 to t = n

• 9 th time distance travelled as

t = 8 to t = 9

• And the time of the object thrown vertically up and reached down is 10 sec.
• Intial velocity of partical is u.

==>

$t = \frac{2u}{g} = 10$

$= \frac{ {u}^{2} }{2g} \\ = \frac{(5 {g}^{2}) }{2g} \\ = \frac{25g}{2}$

=125m.