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A ball is thrown upward with an initial velocity of 100ms^(-1).After how much time will it return?Draw velocity-time graph for the ball and find from the graph (i) the maximum height attained by the ball and (ii) height of the ball after 15 s.Take g=10 ms^(-2) |
Answer» Solution : Here,U =`100ms^(-1)` ,g=`-10 ms^(-2)` At highest point ,v=0 As v=u+gt `therefore=100-10xxt` `therefore` Time taken to reah highest point ,`t=(100)/(10)=10s` The ball will return to the ground at t =20 S Corresponding velocity-time graph of the ball is SHOWN in (i)Maximum height attained the ball =Area of `DeltaAOB=(1)/(2)xx10xx10=500 m` (ii)HEight attained after 15s= Area of `DeltaAOB`+Area of `DeltaBCD` `500+(1)/(2)(15-10)xx(-50)=500-125=375m` |
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