Question

article starts moving along a straight line with a constant acceleration of 4 ms. The ratio of the distance
travelled in the 2nd and 3 seconds is​

Answer 1

Answer:

uy=10sin60=53√m/s

⇒t=2uyg=2×53√10=3√s

Sx=uxt+12axt2

1.15=5×t−12a×t2

1.15=5×3√−32a

3a2=5×1.73−1.15=8.65−1.15

3a2=7.5 

⇒a=153=5m/s2

Explanation:

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