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Grade: 11


A body A is projected upwards with a velocity 98m/s. The second body B is projected with the same initial velocity but after 4s. Both the bodies will meet after

5 years ago

Answers : (4)

Dhulipala Srikanth
45 Points
5 years ago
Shubham Swastik Behera
43 Points
78.4 is the difference of their heights after 10 seconds.
The answer is 12 seconds.
5 years ago
11 Points
							The total time of flight of the body is 10s... So naturally the instant at which they meet will be more than 10s...hence 12sSecondly.... Use s=ut+1/2at2Second body will be t-4i.e ut+1/2at2=u(t-4)+1/2a(t-4)2
3 years ago
14 Points
since both bodies cover equal distance, h1 = h2
here g is negative as the body is projected upwards
therefore, ut -1/2gt^2 = u(t-4) -1/2g(t-4)^2
              ut-1/2gt^2 =  ut – 4u -1/2g(t^2 -8t+16)
               ut-1/2gt^2 =  ut – 4u -1/2gt^2 + 4gt – 8g
                              = -4u – 8g + 4gt
              4u + 8g     =4gt
substituting values, 4*98+8*9.8 = 4*9.8*t
                           4*98+8*9.8/4*9.8 = t
                          4*98/4*9.8 +8*9.8/4*9.8 = t
                              10+ 2 = t
                                   t= 12seconds  
10 months ago
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