If a body projected with a velocity of 19.6 | vedclass

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Question

(B) Given: Initial velocity $u = 19.6 \ m/s$,Maximum height $H = 9.8 \ m$.We know the formula for maximum height is $H = \frac{u^2 \sin^2 	heta}{2g}$

Vedclass · Accepted Answer

$39.2$

Vedclass · Answer

(B) Given: Initial velocity $u = 19.6 \ m/s$,Maximum height $H = 9.8 \ m$.We know the formula for maximum height is $H = \frac{u^2 \sin^2 	heta}{2g}$.Assuming $g = 9.8 \ m/s^2$,we have $9.8 = \frac{(19.6)^2 \sin^2 	heta}{2 	imes 9.8}$.$9.8 = \frac{384.16 \sin^2 	heta}{19.6} \implies 9.8 = 19.6 \sin^2 	heta$.$\sin^2 	heta = \frac{9.8}{19.6} = 0.5$,so $\sin 	heta = \frac{1}{\sqrt{2}}$,which means $	heta = 45^\circ$.The range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$.Since $	heta = 45^\circ$,$\sin(2	heta) = \sin(90^\circ) = 1$.$R = \frac{(19.6)^2}{9.8} = \frac{384.16}{9.8} = 39.2 \ m$.

If a body projected with a velocity of $19.6 \ m/s$ reaches a maximum height of $9.8 \ m$,then the range of the projectile is (Neglect air resistance). (in $m$)

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