MCQ
If we can throw a ball upto a maximum height $H$, the maximum horizontal distance to which we can throw it is
- ✓$2\,H$
- B$\sqrt 2 \,H$
- C$H$
- D$\frac{H}{2}$
✓
Answer
Correct option: A.
$2\,H$
a
$\begin{array}{l}
H = \frac{{{u^2}}}{{2g}} \Rightarrow {u^2} = 2gH\\
For\,{\rm{maximum}}\,horizontal\,{\rm{distance}}\\
{X_{\max }} = \frac{{{u^2}}}{g} = \frac{{2gH}}{g} = 2H
\end{array}$
$\begin{array}{l}
H = \frac{{{u^2}}}{{2g}} \Rightarrow {u^2} = 2gH\\
For\,{\rm{maximum}}\,horizontal\,{\rm{distance}}\\
{X_{\max }} = \frac{{{u^2}}}{g} = \frac{{2gH}}{g} = 2H
\end{array}$
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