Question

Jonhsy was not very strong. (rewrite in affairmative form)​

Answer 1

Here is your answer brooo

Explanation:

Johnsy was not strong

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 :-

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

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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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