Question

a body of mass 2 kg is projected with velocity of 40m/s at an angle 60degree with ground then calculate
(1)range of projectile
(2)vertical component of initial velocity​

Answer 1

We have,

Mass of body, m = 2 kg

Initial velocity of body, u = 40 m/s

Angle of projection, $\sf{\bold{\Theta=60°}}$

Acceleration due to gravity, g = 9.8 m/s²

1.)

Range of projectile is given by,

$\boxed{\sf{R=\frac{u^{2}\sin{(2\Theta)}}{g}}}$

$\implies{\sf{R=\frac{(40)^{2}\sin{(2\times{60°})}}{9.8}}}$

$\implies{\sf{R=\frac{1600\times{\sin{(120°)}}}{9.8}}}$

$\implies{\sf{R=\frac{16000\times{\frac{\sqrt{3}}{2}}}{98}}}$

$\implies{\sf{R=\frac{16000\times{\sqrt{3}}}{98\times{2}}}}$

$\implies{\sf{R=\frac{4000\times{\sqrt{3}}}{49}}}$

$\implies{\boxed{\sf{R=141.39\:m}}}$

2.)

Vertical component of initial velocity is given by,

$\boxed{\sf{u_{y}=u\sin{(\Theta)}}}$

$\implies{\sf{u_{y}=(40)\frac{\sqrt{3}}{2}}}$

$\implies{\sf{u_{y}=20\sqrt{3}}}$

$\implies{\sf{u_{y}=20\times{1.732}}}$

$\implies{\boxed{\sf{u_{y}=34.64\:ms^{-1}}}}$

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