AP EAMCET · PHYSICS · Motion In Two Dimensions
A body is projected at an angle \(\theta\) so that its range is maximum. If \(T\) is the time of flight, then the value of maximum range is (acceleration due to gravity \(=g\) )
- A \(\frac{g^2 T}{2}\)
- B \(\frac{g T}{2}\)
- C \(\frac{g T^2}{2}\)
- D \(\frac{g^2 T^2}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{g T^2}{2}\)
Step-by-step Solution
Detailed explanation
We know that, Range of projectile \[ R=\frac{u^2 \sin 2 \theta}{g} \] As range is maximum, \(\theta=45^{\circ}\) \[ R_{\max }=\frac{u^2 \sin 2 \times 45}{g}=\frac{u^2}{g} \] Flight time of projectile, \[ T=\frac{2 u \sin 45^{\circ}}{g} \]…
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