The range of a projectile of mass 1, kg is 80 | vedclass

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(B) The maximum height $H$ is given by $H = \frac{u_y^2}{2g}$.Given $H = 20\, m$ and $g = 10\, m/s^2$,we have $20 = \frac{u_y^2}{2 	imes 10}$,so $u_y

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$40$

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(B) The maximum height $H$ is given by $H = \frac{u_y^2}{2g}$.Given $H = 20\, m$ and $g = 10\, m/s^2$,we have $20 = \frac{u_y^2}{2 	imes 10}$,so $u_y^2 = 400$,which gives $u_y = 20\, m/s$.The time of flight $T$ is given by $T = \frac{2u_y}{g} = \frac{2 	imes 20}{10} = 4\, s$.The change in momentum $\Delta \vec{P}$ is equal to the impulse provided by the gravitational force,which acts downwards throughout the motion.$\Delta \vec{P} = \vec{F}_{ext} 	imes T = (mg) 	imes T$.Substituting the values: $\Delta \vec{P} = 1\, kg 	imes 10\, m/s^2 	imes 4\, s = 40\, N-s$.

The range of a projectile of mass $1\, kg$ is $80\, m$ while it attains a maximum height of $20\, m$. Its change in momentum during the complete motion is ............ $N-s$.

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