Question

A cracker is thrown into air with a velocity of 10 m/s at an angle of 45${}^o$ with the vertical. When it is at a height of 0.5 m from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of 1 m from the ground? $(g=10m/s^2)$

• A. $4\sqrt{5}m/s$
• B. $2\sqrt{5}m/s$
• C. $4\sqrt{4}m/s$
• D. 10 m/s

Answer 1

The correct option is A $4\sqrt{5}m/s$

Since cracker explodes in mid air it does not encounter any external force so velocity of COM will remain unchanged due to explosion.
Horizontal component $=10\sin {45}^o=\frac{10}{\sqrt{2}}=5\sqrt{2}m/s$
From third equation of motion, at a height of 1 m, vertical component of velocity of center of mass $=\sqrt{{\left(5\sqrt{2}\right)}^2-2\times 10\times 1}=\sqrt{50-20}=\sqrt{30}m/s$

Hence, net velocity $=\sqrt{{\left(5\sqrt{2}\right)}^2+{\left(\sqrt{30}\right)}^2}=\sqrt{80}=4\sqrt{5}m/s$