Discussion
When a body is thrown vertically upwards, from the ground, the time of ascent is \(t_{1}\) and the time of descent is \(t_{2}\) in the absence of air resistance. Then \(t_{1}\) is equal to
Answer: D
\(t_{1}\) is equal to \(t_{2}\).
Without air resistance the motion is symmetric. Going up from speed \(u\) to rest takes \(t_{1} = \frac{u}{g}\), reaching height \(h = \frac{u^{2}}{2g}\). Falling back through the same \(h\) from rest takes \(t_{2} = \sqrt{\frac{2h}{g}} = \frac{u}{g}\).
So the times are equal. The multiples 0.5, 0.25 and 4 would need a different acceleration on the way up and down, which happens only with air resistance (then ascent is shorter than descent).
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