Question

Consider a series of steps as shown. A ball is thrown from 0. Find the minimum speed to directly jump to 5th step

• A. \(5(\sqrt2+1) m/s\)
• B. \(5(\sqrt2) m/s\)
• C. \(5(\sqrt{(\sqrt2+1))} m/s\)
• D. \(6(\sqrt3+1) m/s\)

Answer 1

The Correct Option is C

The correct option is (C): \(5(\sqrt{(\sqrt2+1))} m/s\). The trajectory equation is \(y=x\tan\theta-\frac{gx^2}{2v^2\cos^2\theta}\). The point (2.5, 2.5) must lie on this trajectory, which implies \(1=\tan\theta-\frac{g\times2.5}{2v^2\cos^2\theta}\). Rearranging, we get \(\frac{25}{2v^2\cos^2\theta}=\tan\theta-1\). Further simplification yields \(v^2=\frac{25}{2}\left\{\frac{1+\tan^2\theta}{\tan\theta-1}\right\}\). The minimum velocity \(v_{min}=5\sqrt{\sqrt2+1}\) occurs when \(\tan\theta=\sqrt2+1\).