> Q. The maximum height attained by a projectile is increased by 5%. Keeping the angle of projection constant, what is the percentage increase in horizontal range ? – LIVE ANSWER TODAY

Q. The maximum height attained by a projectile is increased by 5%. Keeping the angle of projection constant, what is the percentage increase in horizontal range ?

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The maximum height attained by a projectile is increased by 5%. Keeping the angle of projection constant, what is the percentage increase in horizontal range ?

Solution:

If $h$ is the maximum height attained by the projectile, then $h =frac{ u ^{2} sin ^{2} theta}{ 2 g }$ or $R =frac{ u ^{2} sin 2 theta }{ g }$

$frac{R}{h}=frac{2 sin theta cos theta}{left(sin ^{2} thetaright) / 2}=4 cot h theta$

$therefore frac{triangle R }{ R }=frac{Delta h }{ h }$

Percentage increase in $R =$ percentage increase in $h = 5 %$.

Percentage increase in $R =$ percentage increase in $h = 5 %$.

$therefore frac{Delta H }{ H }=frac{ 5 }{ 1 0 0 }= 0 . 0 5$

$therefore frac{ 2 Delta u }{ u }= 0 . 0 5$

As, $R =frac{ u ^{2} sin 2 theta }{ g }$

Therefore, $Delta R =frac{2 u Delta u }{ g } times sin 2 theta$

and $frac{Delta R }{ R }=frac{frac{2 u Delta u }{ g } times sin 2 theta}{frac{ u ^{2} sin 2 theta}{ g }}=frac{ 2 Delta u }{ u }= 0 . 0 5$

$therefore %$ increase in horizontal range $=frac{Delta R }{ B } times 100=0.05 times 100=5 %$

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