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  • From a tower of height H, a particle is thrown vertically upwards with a speed of 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

From a tower of height H, a particle is thrown vertically upwards with a speed of 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 are:

Option: 1

2 g H = n2u2


Option: 2

 g H  =  (n - 2)2 u2


Option: 3

2 g H = nu2 (n - 2)


Option: 4

g H = (n - 2)u2


Answers (1)

best_answer

\\ \text{Let } t_{1}\ \text{is time taken to reach the maximum height } \\ v=u+at_{1} \\ 0=u-gt_{1} \ \ \ \ \ \ \ \ \ \ \ \left [ a=-g m/t^{2}, \text{At maximum height, v=0} \right ] \\t_{1}=\frac{u}{g} \ \ \ \ \ -\left ( 1 \right )

\\ \text{Let } t_{2}\ \text{is time taken to hit the ground } \\ S= ut_{2}+\frac{1}{2}at^{2} \\\\ -H=ut_{2}-\frac{1}{2}gt^{2} \ \ \ \ \ -\left ( 2 \right ) \\ \\ \text{According to question},\\t_{2}=nt_{1} =n\frac{u}{g} \ \ \ \ \ -\left ( 3 \right )

\\ \text{Now}, \\ \text{Substitute the value of } t_{2} \ \text{from } \left ( 3 \right )\text{ in equation} \left (2 \right ) \\\\ -H=u\left ( \frac{nu}{g} \right )-\frac{1}{2}g\left ( \frac{nu}{g} \right )^{2} \\ \\ -H=\frac{nu^{2}}{g}-\frac{1}{2}\frac{gn^{2}u^{2}}{g^{2}} \\ \\ -H=\frac{nu^{2}}{g}-\frac{n^{2}u^{2}}{2g} \\\\2gH=nu^{2}\left ( n-2 \right )

 

Posted by

Divya Prakash Singh

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