Question

A particle is projected with a velocity $6\hat{i}+8\hat{j},3\text{m}$ away from a vertical wall(along $\hat{j}$). After striking the wall it lands at a distance of

• A. $3\text{m}$ from the wall
• B. $3.3\text{m}$ from the wall
• C. $5.5\text{m}$ from the wall
• D. $6.6\text{m}$ from the wall

Answer 1

The correct option is D $6.6\text{m}$ from the wall

$T=\frac{2u_y}{a_y}=\frac{2\times 8}{10}=1.6\text{s}$
After collision horizontal velocity gets reversed,
Time taken for particle to reach the wall just befor collision is
$t=\frac{3}{6}=0.5\text{s}.$
After collision particle will travel away from wall with horizontal velocity of $6\text{m/s}$ in time interval of $T-t.$
Hence,
$x=u_x(T-t)=6(1.6-0.5)=6.6\text{m}$