Question

A block of mass $5kg$ is moving horizontally at a speed of $1.5m{s}^{-1}$ . A vertically upward force $5N$ acts on it for $4s$. What will be the distance of the block from the point where the force starts acting?

• A. 2 m
• B. 6 m
• C. 8 m
• D. 10 m

Answer 1

Answer: (D)

Upward acceleration $=\frac{5}{5}=1m/s^2$
Upward distance covered in $4s$
$y=\frac{1}{2}a{t}^2$
$=\frac{1}{2}\times 1\times (4{)}^2=8m$
Horizontal distance covered in $4s$
$x=vt=1.5\times 4=6m$
$\therefore s=\sqrt{x^2+y^2}=\sqrt{6^2+8^2}$
$=\sqrt{36+64}=10m$