Question

A 4 kg projectile is launched from a height of 2 m with an initial speed of 12 m/s. Find its speed when it is at a maximum height of 3 m above the ground.

• A. $\sqrt{31}m{s}^{-1}$
• B. $2\sqrt{31}m{s}^{-1}$
• C. $3\sqrt{31}m{s}^{-1}$
• D. $4\sqrt{31}m{s}^{-1}$

Answer 1

The correct option is B

$2\sqrt{31}m{s}^{-1}$

Given, mass =4 kg

Let initial speed be u =12 m/s

Final speed at a height of 3 m be v

Initial height, $h_1=2m$

Final height, $h_2=3m$

According to law of conserration of energy

Initial kinetic energy + Initial potential energy =Finalkinetic energy+Final potential energy

$\Rightarrow \frac{1}{2}m{u}^2+mg{h}_1=\frac{1}{2}m{v}^2+mg{h}_2$

$\Rightarrow \frac{(12{)}^2}{2}+10\times 2=\frac{1}{2}\times v^2+10\times 3$

$\Rightarrow 72+20=\frac{v^2}{2}+30$

$\Rightarrow \frac{v^2}{2}=62$

$\Rightarrow v=\sqrt{124}m{s}^{-1}=2\sqrt{31}m{s}^{-1}$

Hence, speed at height 3 m is $2\sqrt{31}m{s}^{-1}.$