AP EAMCET · PHYSICS · Motion In One Dimension
A ball is thrown upward from the top of a building at angle of \(30^{\circ}\) to the horizontal with an initial speed of 15 \(\mathrm{ms}^{-1}\). If the ball hits the ground after \(3 \mathrm{~s}\), then the height of building is (acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(30 \mathrm{~m}\)
- B \(12.5 \mathrm{~m}\)
- C \(25.5 \mathrm{~m}\)
- D \(22.5 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(22.5 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\frac{15}{2} \mathrm{~m} / \mathrm{s}\) Time taken by ball to reach the same height of building, \(\mathrm{t}=\frac{2 \mu \sin \theta}{\mathrm{g}}=\frac{2 \times 15 \times \sin 30^{\circ}}{10}=1.5 \mathrm{sec}\) Now, remaining \(1.5 \mathrm{sec}\) will be taken by ball to…
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