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): \(5(\sqrt{(\sqrt2+1))} m/s\)
\(y=x\tan\theta-\frac{gx^2}{2v^2\cos^2\theta}\)
(2.5,2.5) must lie on this
\(⇒1=\tan\theta-\frac{g\times2.5}{2v^2\cos^2\theta}\)
\(⇒ \frac{25}{2v^2\cos^2\theta}=\tan\theta-1\)
\(⇒ v^2=\frac{25}{2}\left\{\frac{1+\tan^2\theta}{\tan\theta-1}\right\}\)
\(⇒ v_{min}=5\sqrt{\sqrt2+1}\)
[ Happens when \(\tan\theta=\sqrt2+1\) ]