Question

Q18
thnx in advance....

Answer 1

We are at height H from the ground (obviously we are standing on a tower of height H) .
we throw a particle vertically upwards with speed 'u'.
taking downward direction as positive .we have at initial point :
initial velocity= -u. ( minus because it is in vertically UPWARD direction.)
acceleration= g (acceleration due to gravity - taken positive- act in downward direction).
Now the time taken(say t) for the particle to reach the highest point is given by:
t= u/g. .....(1) [ by using v=u+at )
(time can never be negative )
further it is given that the total time taken(say T) for the particle to reach the ground is n times the time taken to reach the highest point (which we have calculated in equation (1)).
=> T=nt......(2).
Further,
the total displacement covered by the particle = H ( positive - in downward direction).
total time taken in whole journey = nt = n×(u/g). (using (1) ).
acceleration = g .
then,
H=-u[nu/g]+(1/2)g×(nu/g)^2,
H= -(nu^2)/g + [(nu)^2]/2g,
Taking LCM,
H= {-2(u^2)n + (n^2)(u^2)}/2g,
2gH= -2(u^2)n + (n^2)(u^2),
taking [nu^2] common in RHS..we get..
2gH= n(u^2)[n-2].
Hence option D is correct.
NOTE: When you take downward direction as -ve ,you will get the same result. Feel free to choose.

Answer 2

Let the time taken to reach the highest point be t and the total time be T.

Now, according to the question,
T = nt 

While it goes to the highest point, 
v = u + at 
Here, v = 0
So,

-u = gt (a = g)
Therefore, t = (-u/g)

So, we have : T = (-nu/g)

When it goes down again, 

s = ut + 1/2at²
s = u(-nu/g) + 1/2(g)(-nu/g)(-nu/g)

H = -nu²/g + n²u²/2g 
H/u² = -2n + n²/2g
H/u² = n(n - 2) /2g

2gH = nu²(n - 2) 

Which is option (D).

Happy learning! :D