AP EAMCET · PHYSICS · Motion In Two Dimensions
A ball is projected from ground into the air. At the height of \(5 \mathrm{~m}\), its velocity is \(\mathbf{v}=(5 \hat{i}+5 \hat{j}) \mathrm{ms}^{-1}\). The maximum height reached by the ball is (Acceleration due to gravity \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) )
- A \(8.75 \mathrm{~m}\)
- B \(5.50 \mathrm{~m}\)
- C \(6.25 \mathrm{~m}\)
- D \(10 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(6.25 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The motion of projectile is shown below To calculate height attained we consider only vertical motion. Here, \(u=5 \mathrm{~m} / \mathrm{s} ; a=-10 \mathrm{~m} / \mathrm{s}^2\) \(v=0\) From \(v^2-u^2=2 a s\) We have \(0-5^2=2(-10) s\)…
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