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InterviewSolution
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A stone is thrown vertically upwards with an initial velocity of 40 m s^(-1). Taking g = 10 m s^(-2), draw the velocity-time graph of the motion of stone till it comes back on the ground. (i) Use graph to find the maximum height reached by the stone. (ii) What is the net displacement and total distance covered by the stone ? |
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Answer» Solution :Given u = 40 `m s^(-1)` g =10 m `s^(-2)`. As the stone rises up, the velocity decreases at the rate of 10 m `s^(-2)`. When the velocity becomes zero, the stone is at its highest position. Then it begins to fall and its velocity increases at a rate of 10 m `s^(-2)`. The velocity of stone at DIFFERENT instants is shown in the following table (the upward direction is taken positive).  Fig 2.29 SHOWS the velocity - time graph .  Maximum height reached by the stone = Area of `Delta OAB` `= (1)/(2) OBxx OA ` `= (1)/(2) xx4s xx 40 m s^(-1)= 80 `m. (ii) Net displacement = Area of `Delta `OAB- Area of `Delta BDC` `= (1)/(2) OBxxOA - (1)/(2) BD xx DC` `= ((1)/(2) xx 4S xx 40 m s^(-1)) - ((1)/(2) xx 4s xx 40 m s^(-1))=0` Total distance covered = Area of `Delta` OAB + Area of `Delta` BDC = 80 m + 80 m = 160 m |
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