Question

pls koi yeh question kardo...........​

Answer 1

SOLUTION

Given:

there is maximum possible range of 400metre.

So, angle of projection=45°

$= > \frac{ {u}^{2}sin2 \theta}{g} = 400 \\ = > {u}^{2} = 4000 \\ \\ = > hmax = \frac{4y {}^{2} }{2g} = \frac{ {u}^{2}sin {}^{2} \theta }{2g} =\frac{4000 \times \frac{1}{2} }{20} (sin45 = \frac{1}{ \sqrt{2} } ) \\ = > 100 \\ = > hmax \: will \: be \: at \: half \: the \: horizontal \: distance = \frac{r}{2} \\ = > \frac{400}{2} = 200 = x \\ = > so \: the \: co - ordinate \: = (200 \: or \: 100)$

Option (b)✓

hope it helps ✔️❣️