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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 ​

Answer» <html><body><p>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 <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> = 10....H=0+1/2x144H=72m 1...height of tower is 72 m2..distance in 12 sec = 72 <a href="https://interviewquestions.tuteehub.com/tag/m3-1082064" style="font-weight:bold;" target="_blank" title="Click to know more about M3">M3</a>..again use the formula <a href="https://interviewquestions.tuteehub.com/tag/instead-516622" style="font-weight:bold;" target="_blank" title="Click to know more about INSTEAD">INSTEAD</a> of 12 take 10....H=0+1/2x100H=50 mFor 10 sec..50 m4.....50 m..<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> ......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..</p></body></html>


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