An object of mass 10kg is projected from ground with speed 40m-s 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-

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Toppr Admin · Accepted Answer

According to intial data we have-v-x-40-cos60-160- -160-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-

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 $$g=9.8m/s^2$$]

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 = $$m(v_y+v_x)4m(v_y+v_x)$$
$$=m(40\sin60-9.8+40\cos60)$$
$$=m(40\sin60+4\cos60)$$
$$=m(9.8)$$
$$=9.8$$

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 $$g=9.8m/s^2$$]

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 = $$m(v_y+v_x)4m(v_y+v_x)$$$$=m(40\sin60-9.8+40\cos60)$$$$=m(40\sin60+4\cos60)$$$$=m(9.8)$$$$=9.8$$

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 = $$m(v_y+v_x)4m(v_y+v_x)$$
$$=m(40\sin60-9.8+40\cos60)$$
$$=m(40\sin60+4\cos60)$$
$$=m(9.8)$$
$$=9.8$$