Question
A body is projected vertically upwards with a velocity
of 98 m/s. The time interval during which it is at a height greater than 392 m is: (Take g = 9.8 m/s²)Solution
Maximum height H = u²/2g = 98²/(2 × 9.8) = 490 m Using the equation h = ut − (1/2)gt², set h = 392 Solve 98t − 4.9t² = 392 ⇒ 4.9t² − 98t + 392 = 0 Solving the quadratic gives roots t₁ = 4s, t₂ = 12s So, height > 392 m for time interval t = 12 − 4 = 8s.
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A Trial Balance contains the balance of:
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