Question
a body is p rojected vertically upwards with speed 20m/s .find the distance travelled by it during the last second of its upward journey(takeg=10m/s2
Joshi sir comment
last sec. of upward journey means first sec. of free fall from the top so s=0(1) + 1/2(10)(1)2 = 5 meter, here g is taken 10m/s2
Read 1 Solution.
s= u - a/2 (2t-1)= u- g/2 ( 2u/g-1) here last sec is used there4 t=u/g will be used , remember t=1 will be taken if first sec is asked & here last is asked therefore t=1 will not be used ( u=20, g=10 on putting these values u vyl get d ans)
SARIKA 12 year ago
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