Question

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is
A-\[2gH=n{{u}^{2}}(n-2)\]
B-\[gH=(n-2){{u}^{2}}\]
C-\[2gH={{n}^{2}}{{u}^{2}}\]
D-\[gH={{(n-2)}^{2}}{{u}^{2}}\]

Answer 1

Hint: the motion is in one dimension. No external force is acting on the system and the only force which acts is the gravitational force and the body moves both upside and down under the influence of gravitational force. Acceleration due to gravity is taken to be g.

Complete step by step answer:
The particle is thrown vertically upwards from the top of the tower and the height of the tower is H.
Initial velocity=u
Final velocity=0
Acceleration=-g
Using v=u+at
0=u-gt
\[{{t}_{1}}=\dfrac{u}{g}\]--------------(1)
If \[{{t}_{2}}\] be the time taken to hit the ground then:
\[-H=u{{t}_{2}}-\dfrac{1}{2}\times gt_{2}^{2}\]
But it is given in the question that \[{{t}_{2}}=n{{t}_{1}}\]
Putting this in above equation we get
\[\begin{align}
  & -H=u\dfrac{nu}{g}-\dfrac{1}{2}\times g\dfrac{{{n}^{2}}{{u}^{2}}}{{{g}^{2}}} \\
 & 2gH=n{{u}^{2}}(n-2) \\
\end{align}\]

So, the correct answer is “Option A”.

Additional Information:
 Also when a body is thrown upwards then always at the topmost point the body takes a momentarily pause and when it starts to come back down towards the earth, in this case, the initial velocity is taken as zero and body strikes the earth back with some velocity.

Note:
This problem we have to keep in mind that any motion in an upward direction, the acceleration due to gravity is negative and in downward motion acceleration due to gravity is taken positively. All other values to be taken in standard SI units.