1.

2) r8 m(3) 15.6 m(4) 39.2 mTwo bodies are thrown vertically úpward, with the same initial velocity of 98 m/s but 4 sec apartHow long after the first one is thrown when they meet ?(2) 11 sec(1) 10 sec(3) 12 sec(4) 13 sec

Answer»

When they meet, their displacements will be same. Therefore ,

s= ut-(1/2)gt^2=u(t-4) -(1/2)g(t-4)^2

Simplifying and making t subject, we get

t=(u+2g)/g. OR

Substituting the values,

t=(98+19.6)/9.8 OR

t=12s after the first body was thrown.

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