Question

a body is thrown up with a velocity 29.23m/s. Distance travelled in last second of upward motion is

Answer 1

Assuming the body will fall and terminal velocity is not reached, find out the distance it will travel up by using v² = u²+2as where v² is 0 and a is -9.81 Then use the the same distance again in the same equation to find the ending velocity v v² = u²+2as where u² is 0 and a is 9.81 Note :- Dont mess up the signs of acceleration


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Answer 2

applying the law,v²=u²-2a(-h). h is negative becuz the object is moving against the gravity. and here, u=29.23m/sand v=0m/s.so we get h=43.59metres(total distance)then time taken to cover the distance=43.59/29.23=1.49seconds. and distance covered by the body in the last second=43.59-14.32=29.27m.