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InterviewSolution
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A particle released from top of a tower reaches ground after 12 sec calcute the following 1.Height of tower 2.distance travelled by tower in 12th sec 3 . Distance travelled in 10 sec 4. Distance travelled in 10th sec 5. distance travelled in 5th , 7th, and 9th sec |
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Answer» The formula here will be used as. s= ut +1/2 at squareintial speed of stone = 0m/stime taken by the particle =12 secheight of the tower.. H=ut +1/2 at square H=0x12+1/2x10x12x12 ......Gravitation CONSTANT = 10....H=0+1/2x144H=72m 1...height of tower is 72 m2..distance in 12 sec = 72 M3..again use the formula INSTEAD of 12 take 10....H=0+1/2x100H=50 mFor 10 sec..50 m4.....50 m..5 ......In 5 sec... H=0+1/2x25 H=12.5 m ....In 7 sec... H=0+1/2x49 H=24.5 m........In 9 sec.... H=0+1/2x81 H=40.5 m.. |
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