a ball is dropped from a height if it takes 0.2 seconds to cross the last 6 m before hitting the ground then the height from which it was dropped (g=9.8m/s)
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using
s=ut +1/2 at2
h=6+ 1/2×9.8×(0.2)²
h=6+9.8×0.02
h=6+0.196=6.196m
s=ut +1/2 at2
h=6+ 1/2×9.8×(0.2)²
h=6+9.8×0.02
h=6+0.196=6.196m
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