TS EAMCET · Physics · Motion In Two Dimensions
A ball projected at an angle of \(45^{\circ}\) with the horizontal crosses two points at equal heights separated by a distance at times 2 s and 8 s respectively. The horizontal distance between the two points is
(Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A 300 m
- B 400 m
- C 500 m
- D 600 m
Answer & Solution
Correct Answer
(A) 300 m
Step-by-step Solution
Detailed explanation
\( u \sin\theta = \frac{g(t_1 + t_2)}{2} \) \( u \sin 45^{\circ} = \frac{10(2 + 8)}{2} = 50 \text{ ms}^{-1} \) Since \( \theta = 45^{\circ} \), \( u \cos\theta = u \sin\theta = 50 \text{ ms}^{-1} \) \( \Delta x = u \cos\theta (t_2 - t_1) \)…
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