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After a ball has been falling under gravity for 5 seconds it passes through a piece of glass and loses half velocity. If it now reaches the ground in 1 seconds find the height of the glass above the ground. A particle is projected vertically upward from ground level with a speed of 50 m/s. For how long will it be more than 70 m above the ground?
Dear Shubham,
v = v0 + gt = 0 + (9.8m/s²)(5.0s) = 49m/s After breaking through the glass, its speed has been reduced to roughly half this: 25m/s. If it takes 1.0s to reach the ground with that speed, then the height of the glass was: Δy = v0t + 0.5gt² = (25m/s)(1.0s) + 0.5(9.8m/s²)(1.0s)² = 30m (rounded)
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