Physics - NLM Question with Solution | TestHub

Let the thread make angle $\theta$ with the ground and the body's speed be $v$; the engine winds thread at $v_0$, so $v\cos \theta =v_0$. The body lifts off when the thread's vertical pull balances weight: $T\sin \theta =Mg$, and horizontally $T\cos \theta =M\frac{dv}{dt}\Rightarrow \frac{dv}{dt}=g\cot \theta$.

Differentiate $v\cos \theta =v_0$: $\frac{dv}{dt}\cos \theta =v\sin \theta \frac{d\theta }{dt}$. With $x=H\cot \theta$, this gives $v_0^2=gH{\cot }^4\theta$.

$x=H\cot \theta =H{\left(\frac{v_0^2}{gH}\right)}^{1/4}=0.1{\left(\frac{16}{10\times 0.1}\right)}^{1/4}=0.1\times 16^{1/4}=0.2\mathrm{m}$.