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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2014 · 6 April · Physics · Motion in One Dimension

From a tower of height H, a particle is thrown vertically upwards...

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 :From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to

Options

  1. A2 g H = n2u2
  2. Bg H = (n - 2)2u2
  3. C2g H = nu2(n - 2)
  4. Dg H = (n - 2)u2

Correct answer

C. 2g H = nu2(n - 2)

Step-by-step solution

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to

ntAB=tAB+tBC+tCD{\text{nt}}_{\text{AB}} = {\text{t}}_{\text{AB}} + {\text{t}}_{\text{BC}} + {\text{t}}_{\text{CD}}

\left(\text{n} - 1\right) {\text{t}}_{\text{AB}} = {\text{t}}_{\text{BC}} + {\text{t}}_{\text{CD}} ....(1)<br/><br/>(n−1)ug=ug+tCD<br /><br />\left(\text{n} - 1\right) \dfrac{\text{u}}{\text{g}} = \dfrac{\text{u}}{\text{g}} + {\text{t}}_{\text{CD}}

∴(n−2)ug=tCD∴ \left(\text{n} - 2\right) \dfrac{\text{u}}{\text{g}} = {\text{t}}_{\text{CD}} ...(2)

H+h=12gtBD2\text{H} + \text{h} = \dfrac{1}{2} {\text{gt}}_{\text{BD}}^{2}

H+u22g=12g[(n−1)ug]2\text{H} + \dfrac{{\text{u}}^{2}}{2 \text{g}} = \dfrac{1}{2} \text{g} {\left[\dfrac{\left(\text{n} - 1\right) \text{u}}{\text{g}}\right]}^{2} ...from (1)

∴H+u22g=(n−1)2u22g∴ \text{H} + \dfrac{{\text{u}}^{2}}{2 \text{g}} = \dfrac{{\left(\text{n} - 1\right)}^{2} {\text{u}}^{2}}{2 \text{g}}

∴H=u22g[(n−1)2−1]∴ \text{H} = \dfrac{{\text{u}}^{2}}{2 \text{g}} \left[{\left(\text{n} - 1\right)}^{2} - 1\right]

2g H = nu2(n - 2)
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