Question

A particle is projected from ground with velocity 40√2 meter per second at 45degree find velocity and displacement of the particle after 2second

Answer 1

Given,

$u=40\sqrt2\ \mathrm{m\ s^{-1}}\\\\\theta=45^{\circ}$

Let $g=10\ \mathrm {m\ s^{-2}}.$

Velocity of a body projected with a velocity u making an angle θ with the horizontal at a time t is given by,

$v=\sqrt{u^2+g^2t^2-2ugt\sin\theta}$

So, at $t=2,$

$v=\sqrt{(40\sqrt2)^2+100\cdot4-2\cdot10\cdot 2\cdot40\sqrt{2}\sin 45^{\circ}}\\\\v=\sqrt{3200+400-1600}\\\\v=\mathbf {20\sqrt{5}\ m\ s^{-1}}$

Displacement is given by,

$s=\sqrt {\dfrac {1}{4}g^2t^4-ug\sin\theta t^3+u^2t^2}$

So, at $t=2,$

$s=\sqrt {\dfrac {1}{4}\cdot 100\cdot 16-40\sqrt2 \cdot10\cdot\sin 45^{\circ}\cdot 8+(40\sqrt2)^2\cdot 4}\\\\\\s=\sqrt{400-3200+12800}\\\\\\s=\mathbf {100\ m}$