Question

5. A stone is released from a certain height, h. In the last sec of its motion, it moves 9 times the
distance travelled in the first second of its journey. Find the value of h.​

Answer 1

Answer:

traveling distance in 1st sec

Sn = u + a(n-1/2)

S1 = 0+ g (1-1/2)

S1 = g/2

u+g (n - 1/2) = 9. g/2

u + ng = 9.g/2 +g/2

u + ng = 10. g/2

u+ ng = 5g

u = 5g - ng

H = ut + 1/2 gt^2

= (5g-ng).n + 1/2 g n^2

= 5gn - gn^2 + 1/2 g.n^2

= 5gn + gn^2 (-1 + 1/2)

=(10 ng - gn^2)/2

= ng ( 10 - n)/2

=10 n (10-n)/2

=5n (10-n)