WBJEE · Physics · Motion In Two Dimensions
From the top of a tower, \(80 \mathrm{m}\) high from the ground, a stone is thrown in the horizontal direction with a velocity of \(8 \mathrm{ms}^{-1}\). The stone reaches the ground after a time \(t^{\prime}\) and falls at a distance of " \(\mathrm{d}\) ' from the foot of the tower. Assuming \(g=10 \mathrm{m} / \mathrm{s}^{2}\), the time \(t\) and distance d are given respectively by
- A \(6 \mathrm{s}, 64 \mathrm{m}\)
- B \(6 \mathrm{s},48 \mathrm{m}\)
- C \(4 \mathrm{s}, 32 \mathrm{m}\)
- D \(4 \mathrm{s}, 16 \mathrm{m}\)
Answer & Solution
Correct Answer
(C) \(4 \mathrm{s}, 32 \mathrm{m}\)
Step-by-step Solution
Detailed explanation
Given, \(h=80 \mathrm{m}, v=8 \mathrm{ms}^{-1}\) and \(\mathrm{g}=10 \mathrm{ms}^{-2}\) Here, \[ \begin{aligned} h &=\frac{1}{2} g t^{2} \\ 80 &=\frac{1}{2} \times 10 \times t^{2} \\ t &=4 \mathrm{s} \end{aligned} \] and \(\quad x=v t=8 \times 4=32 \mathrm{m}\)
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