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

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 = n2u2 (B)  gH = (n-2)2 u2 (C) 2gH = nu2 (n-2) (D)  gH = (n-2)u2

Tardigrade Admin · Accepted Answer

Time taken to reach highest point is t1=(u/g) Speed on reaching ground =√u2+2gh Now, v=u+at ⇒ √u2+2 g h=-u+gt ⇒ t=(u+√u2+2 g H/g)=(nu/g) ⇒ 2gH=n(n-2) u2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0732013bc8f1f2e3d29df276d666f961-.png" />

Q. 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

$2 g H = n^{2} u^{2}$

$g H = (n - 2)^{2} u^{2}$

$2 g H = n u^{2} (n - 2)$

$g H = (n - 2) u^{2}$

Time taken to reach highest point is $t_{1} = \frac{u}{g}$
Speed on reaching ground $= u^{2} + 2 g h$
Now, $v = u + a t$
$\Rightarrow u^{2} + 2 g h = - u + g t$
$\Rightarrow t = \frac{u + u ^{2} + 2 g H}{g} = \frac{n u}{g}$
$\Rightarrow 2 g H = n (n - 2) u^{2}$

Q. 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

2gH=n2u2

gH=(n−2)2u2

2gH=nu2(n−2)

gH=(n−2)u2