Question

An object of mass $$10kg$$ is projected from ground with speed $$40m/s$$ at an angle $$60^0$$ with horizontal the rate of change of momentum of object one second after projection in SI unit is [Take $$g=9.8m/s^2$$]

Solution
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According to intial data we have
$$v_x=40\cos60$$ (const)
$$v_y=40\sin60-(9.8)(1)$$
$$v_y=40\cdot\frac{\sqrt{3}}{2}(9.8)(1)$$
Change in momentum = $$m(v_y+v_x)4m(v_y+v_x)$$
$$=m(40\sin60-9.8+40\cos60)$$
$$=m(40\sin60+4\cos60)$$
$$=m(9.8)$$
$$=9.8$$

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