A 0.5 kg ball is thrown up with an initial speed 14 m/s and reaches maximum height of 8.0 m how much energy dissipated by air drag acting on the ball during the ascent
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If there is no air drag then maximum height,
H=u^2/2g=14×14/2×9.8=10m
But due to air drag ball reaches up to height 8m only.
So, loss in energy
=mg(10−8)
=0.5×9.8×2=9.8J
H=u^2/2g=14×14/2×9.8=10m
But due to air drag ball reaches up to height 8m only.
So, loss in energy
=mg(10−8)
=0.5×9.8×2=9.8J
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131
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