TS EAMCET · Physics · Motion In Two Dimensions
A small ball is thrown at an angle \(45^{\circ}\) to the horizontal with an initial velocity of \(2 \sqrt{2} \mathrm{~m} / \mathrm{s}\). The magnitude of mean velocity averaged over the first \(2 \mathrm{~s}\) is [talse, acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\) ]
- A \(7.0 \mathrm{~m} / \mathrm{s}\)
- B \(8.2 \mathrm{~m} / \mathrm{s}\)
- C \(7.8 \mathrm{~m} / \mathrm{s}\)
- D \(9 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(8.2 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Horizontal displacement in 2 seconds, \[ x=u_x \times t=2 \sqrt{2} \cos 45^{\circ} \times 2=4 \mathrm{~m} \] Vertical displacement in 2 seconds,…
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