WBJEE · Maths · Basic of Mathematics
A particle is projected vertically upwards. If it has to stay above the ground for 12 seconds, then
- A velocity of projection is \(192 \mathrm{ft} / \mathrm{sec}\)
- B greatest height attained is \(600 \mathrm{ft}\)
- C velocity of projection is \(196 \mathrm{ft} / \mathrm{sec}\)
- D greatest height attained is \(576 \mathrm{ft}\)
Answer & Solution
Correct Answer
(D) greatest height attained is \(576 \mathrm{ft}\)
Step-by-step Solution
Detailed explanation
Hint: \(V=u-g t\) at \(t=6\) \(u-g t=0\) \(\Rightarrow u=6 g=192 f t /\) sec \(\quad\left(g=32 f t / \sec ^{2}\right) \ldots \ldots . .(\mathrm{i})\) \(x=u t=\frac{1}{2} g t^{2}\) \(=192.6-\frac{1}{2} 32.6^{2}\) \(=576 \mathrm{ft}\)
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