WBJEE · Physics · Motion In Two Dimensions
A cricket ball thrown across a field is at heights \(h_{1}\) and \(h_{2}\) from the point of projection at times \(t_{1}\) and \(t_{2}\) respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is
- A \(\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}\)
- B \(\frac{h_{1} t_{1}^{2}+h_{2} t_{2}^{2}}{h_{2} t_{1}+h_{1} t_{2}}\)
- C \(\frac{h_{1} t_{2}^{2}+h_{2} t_{1}^{2}}{h_{1} t_{2}+h_{2} t_{1}}\)
- D \(\frac{h t_{1}^{2}-h t_{2}^{2}}{h_{1}-h_{2} t_{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}\)
Step-by-step Solution
Detailed explanation
For vertically moment (for \(\left.h_{1}\right)\) or \(\begin{array}{l} h_{1}=u \sin \theta t_{1}-\frac{1}{2} g t_{1}^{2} \\ t_{1}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{u \sin \theta} \end{array}\) \(h_{2}=u \sin \theta t_{2}-\frac{1}{2} g t_{2}^{2}\) or…
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