AP EAMCET · PHYSICS · Motion In Two Dimensions
Two towers \(A\) and \(B\), each of height \(20 \mathrm{~m}\) are situated a distance \(200 \mathrm{~m}\) apart. A body thrown horizontally from the top of the tower \(A\) with a velocity \(20 \mathrm{~ms}^{-1}\) towards the tower \(B\) hits the ground at point \(P\) and another body thrown horizontally from the top of tower \(B\) with a velocity \(30 \mathrm{~ms}^{-1}\) towards the tower \(A\) hits the ground at point \(Q\). If a car starting from rest from \(P\) reaches \(Q\) in 10 seconds, then the acceleration of the car is \(\left(\right.\) acceleration due to gravity \(\left.=10 \mathrm{~ms}^{-2}\right)\)
- A \(1 \mathrm{~ms}^{-2}\)
- B \(2 \mathrm{~ms}^{-2}\)
- C \(3 \mathrm{~ms}^{-2}\)
- D \(4 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Given, height of both towers is same, \(h_1=h_2=h\). Time of flight, \(t=\sqrt{\frac{2 h}{g}}\) will be same \( t=\sqrt{\frac{2 \times 20}{10}}=\sqrt{4}=2 \mathrm{~s} \) \(\Rightarrow\) Displacement in horizontal direction from tower \(A\) to point \(P=u_A t\)…
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