Question

The horizontal range of a projectile fired at an angle of 15° is 50m. If it fired with the same speed at angle of 45°, It's range is ?​

Answer 1

${\underline{\sf{\purple{Given}}}:-}$

$\:\:\:\:\bullet\:\:\:\:\sf{ The \ horizontal \ range \ of \ a \ projectile \ fired \ at \ an \ angle \ of \ 15^{\circ} \ is \ 50m.}$

${\underline{\sf{\purple{ToFind}}}:-}$

$\:\:\:\:\bullet\:\:\:\:\sf{If \ it \ fired \ with \ the \ same \ speed \ at \ angle \ of \ 45^{\circ}, \ It's \ range \ is \ \?}$

${\underline{\sf{\purple{Solution}}}:-}$

We know,

$\:\:\:\bullet\:\:\sf{Range\:(R)=\dfrac{u^{2}sin2\theta}{g}}$

$\:\:\:\bullet\:\:\sf{\theta = Angle \: of \: projection}$

$\:\:\:\bullet\:\:\sf{\theta = 15^{\circ}}$

$\:\:\:\bullet\:\:\sf{R = 50m}$

Putting the values in the range formula

$\\\dashrightarrow\:\sf{R=50m=\dfrac{u^{2}sin(2\times15^{\circ})}{g}}$

$\dashrightarrow\:\sf{50\times g = u^{2}sin30^{\circ}=u^{2}\dfrac{1}{2}}$

$\dashrightarrow\:\sf{50\times g \times 2 = u^{2}}$

$\dashrightarrow\:\sf{u^{2} = 50\times9.8\times2=100\times9.8=980}$

$\dashrightarrow\:\sf{u = \sqrt{980} = \sqrt{49\times 20}=7\times 2\sqrt{5}m/s}$

$\dashrightarrow\:\sf{=14\times 22.23m/s = 31.304m/s}$

For θ = 45° ;

$\\\dashrightarrow\:\sf{R = \dfrac{u^{2}sin2\times 45^{\circ}}{g} = \dfrac{u^{2}}{g}}$

$\dashrightarrow\:\sf{R = \dfrac{(14\sqrt{5})^{2}}{g} = \dfrac{14\times14\times5}{9.8}}$

$\dashrightarrow\:{\boxed{\sf{\red{100m}}}}$

Answer 2

$\begin{gathered}\begin{gathered}\sf Given -  \begin{cases} &\sf Horizontal \:Range\:(R_1)\:= \frak{50\:m}\\ & \sf Angle \:of \:projection \:theta_1\:=\:\frak{15°}\\ & \sf Angle \:of\: projection\:theta_2\:=\:\frak{45°} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\sf To \:  Find -   \begin{cases} &\sf New\:Range\:(R_2) \end{cases}\end{gathered}\end{gathered}$

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• Here,It is given that the horizontal range of a projectile fired at an angle of 15° is 50 m. So,the horizontal range of a projectile is defined as the horizontal displacement. Mathematically, it can be written as :-

$\large\pink{\boxed{\sf R=\frac{u^2sin2\theta}{g}}}$

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

Now, according to the first condition we have to find initial speed of the projectile. Where, we know the value of g ( 9.8 meter per second square) . So, the formula will be :-

$\purple{\boxed{\sf R_1=\frac{u^2sin2\theta_1}{g}}}$

❍Now, Substitute the given values :-

$\:\:\: \:\:\:\:\: :\implies{\sf 50=\dfrac{u^2sin2×15°}{9.8}}$

$\: \:\: \:\: \:\:\: :\implies{\sf 50=\dfrac{u^2×sin30°}{9.8} }$

$\:\:\: \:\: \:\: \: :\implies{\sf u^2=50×9.8×2}$

$\:\:\: \:\: \:\: \: :\implies{\sf u^2=980m/s}$

$\:\:\: \:\:\:\:\::\implies{\underline{\boxed{\frak{\purple{u= 31.30\;m/s}}}}}\;\bigstar$

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Again, according to the second condition, if it is fired with same speed at an angle of 45°. Its range will be :-

$\red{\boxed{\sf R_2=\frac{u^2sin2\theta_2}{g}}}$

$\:\:\: \:\:\:\:\::\implies{\sf R_2=\dfrac{(31.30)^2×sin2×45°}{9.8} }$

$\:\:\: \:\:\:\:\::\implies{\sf R_2=\dfrac{980×sin90°}{9.8} }$

$\:\:\: \:\:\:\:\::\implies{\sf R_2=\dfrac{980×1}{9.8} }$

$\:\:\: \:\:\:\:\::\implies{\underline{\boxed{\frak{\red{R_2= 100\;m}}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence, \;the\;new\; range \;is\;\bf{ 100\;m}.}}}$

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