TS EAMCET · Physics · Motion In Two Dimensions
A helicopter flying horizontally with a velocity of 288 kmph drops a bomb. If the line joining the point of dropping the bomb, and the point where bomb hits the ground makes an angle \(45^{\circ}\) with the horizontal, then the height at which the bomb was dropped is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A 1320 m
- B 1280 m
- C 320 m
- D 640 m
Answer & Solution
Correct Answer
(B) 1280 m
Step-by-step Solution
Detailed explanation
\(u = 288 \text{ kmph} = 288 \times \frac{5}{18} \text{ m/s} = 80 \text{ m/s}\) \(\tan 45^{\circ} = \frac{H}{R} \Rightarrow H = R\) \(H = \frac{1}{2}gt^2\) and \(R = ut \) \(\frac{1}{2}gt^2 = ut \Rightarrow t = \frac{2u}{g}\)…
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