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Normal 0false false falseEN-US X-NONE X-NONE/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:Times New Roman;mso-bidi-theme-font:minor-bidi;}A stone is thrown vertically upwards. When particle is at a height half of its maximum height, its speed is 10m/s; then the maximum height attained by the particle is: [g = 10m/s2](a) 8 m(b) 10 m(c) 15 m(d) 20 m

Aditi Chauhan , 11 Years ago
Grade 10
anser 2 Answers
Jitender Pal

Last Activity: 11 Years ago

(b)
Let h be maximum height to be attained by stone.
Velocity with which particle is thrown
    u= √2gh
    v^2=u^2+2gh
    when H=h/2,u= √2gh,v=10 m/s   g=-g
    v^2-u^2=  2gH 

2gh-100=  2g h/2
        2 g  h/2  =100 

   h =10 m.

Arpit Aggarwal

Last Activity: 5 Years ago

V^2 - u^2 = 2aS
10^2 - u^2 = -2gH/2
-gH = 100 - u^2
-gu^2 / (2g) = 100 - u^2
u^2 / 2 = u^2 - 100
100 = u^2 / 2
u^2 = 200
H = u^2 / (2g)
= 200 / (2 × 10)
= 10 m
Maximum height attained by it is 10 m
 

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