Question

It is given that the ball hits a wall at a horizontal distance $15m$ at a point $5m$ above the point of projection. If a second wall of height $($$5-dx$$)$ $m$ with $dx\to 0$ has to be constructed such that the ball just misses to hit this wall, horizontal distance of this wall from the point of projection will be

• A. $7.5$ $m$
• B. $12$ $m$
• C. $10.5$ $m$
• D. $6.5$ $m$

Answer 1

The correct option is A $7.5$ $m$

In a projectile the particle follow a specific path which call the trajectory of the particle , the equation of the trajectory as follows -

$y=x\tan \theta -\frac{g{x}^2}{2{\left(u\cos \theta \right)}^2}$

where x is the horizontal displacement and y is the vertical displacement of the particle at any point of the time during projectile.
now put the given values in the equation to get the velocity
$⟹5=15tan\left({45}^{\circ }\right)-\frac{10{\times 15}^2}{2{\left(ucos\left({45}^{\circ }\right)\right)}^2}$

$⟹u=15m/s$

As we know that the particle passes the $5m$ twice so by again putting all the values in the trajectory equation you will get the another distance,

So,
$⟹5=x\times tan\left({45}^{\circ }\right)-\frac{10{\times x}^2}{2{(15\times cos\left({45}^{\circ }\right))}^2}$

$⟹x=7.5m$

So the Horizontal distance of the new wall is $7.5m$.