TS EAMCET · Physics · Motion In Two Dimensions
A small object is thrown at an angle \(45^{\circ}\) to the horizontal with an initial velocity \(\mathbf{v}_0\). The velocity is averaged for first \(\sqrt{2} \mathrm{~s}\) and the magnitude of average velocity comes out to be same as that of initial velocity, i.e. \(\left|\mathbf{v}_{\mathrm{o}}\right|\). The magnitude \(\left|\mathbf{v}_0\right|\) will be (take, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(3 \mathrm{~m} / \mathrm{s}\)
- B \(3 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
- C \(4 \mathrm{~m} / \mathrm{s}\)
- D \(5 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(D) \(5 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Let object is at \(B(x, y)\) after \(t=\sqrt{2} \mathrm{~s}\) Then, \(x=u_x \times t=v_0 \cos 45^{\circ} \times \sqrt{2}=v_0\) and \(y=u_y t-\frac{1}{2} a_y t^2=v_0\left(\sin 45^{\circ}\right) \sqrt{2}-\frac{1}{2}(10) \times(\sqrt{2})^2\) \( =\left(v_0-10\right) \mathrm{m} \)…
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