a body is thrown vertically upward and takes 5 secs to reach maximum height the distance travelled by the body will be same in 1)1st and 10th second2)2nd and 8th second3)4th and 6th second 4)both 2 and 3
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the answer is 1) 1st and 10th second
method:
using g = 10 m/sec^2 :
Vo = g*t = 10*5 = 50 m/sec
from t = 0 to t = 1 sec
h = ((Vo*1-5*1^2)-0) = (50-5) = 45 m (up)
from t = 9 to t = 10 sec
h = ((Vo*10-5*10^2)-(Vo*9-5*9^2) = (0-450+405) = -45 m (down)
method:
using g = 10 m/sec^2 :
Vo = g*t = 10*5 = 50 m/sec
from t = 0 to t = 1 sec
h = ((Vo*1-5*1^2)-0) = (50-5) = 45 m (up)
from t = 9 to t = 10 sec
h = ((Vo*10-5*10^2)-(Vo*9-5*9^2) = (0-450+405) = -45 m (down)
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