Question

The horizontal range of a projectile fired at an angle of 15 degree is 50 m .If it is fired at the same speed at an angle of 45 degree.It range will be

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf R_2=100m}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline{\underline{\sf Given:}}$

• Horizontal Range ($\sf{R_1}$)=50m

• Angle of projection ($\sf{\theta_1}$)=15°

• Angle of projection ($\sf{\theta_2}$=45°

$\large\underline{\underline{\sf To\:Find:}}$

• New Range ($\sf{R_2}$)=?

_______________________________________

$\huge\underline{\underline{\sf Range(R_1):}}$

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

$\large\implies{\sf 50=\frac{u^2sin2×15°}{10}}$

$\large\implies{\sf 50=\frac{u^2×sin30°}{10} }$

$\large\implies{\sf 50=\frac{u^2×1}{10×2}}$

$\large\implies{\sf u^2=50×10×2}$

$\large\implies{\sf u^2=500×2}$

$\large\implies{\sf u^2=1000m/s}$

According to Question next projectile is fired with the same Velocity therefore Velocity (u) will be :-

$\sf{u^2=1000m/s}$ for next projectile .

Now for next projection :

$\huge\underline{\underline{\sf Range(R_2):}}$

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

$\large\implies{\sf R_2=\frac{1000×sin2×45°}{10} }$

$\large\implies{\sf R_2=\frac{1000×sin90°}{10} }$

$\large\implies{\sf R_2=\frac{1000×1}{10} }$

$\large\implies{\sf R_2=100m}$

$\large\red{\boxed{\sf R_2=100m}}$

Hence ,

Range ($\sf{R_2}$) for new projectile is 100m

$\large{♡}$MᴏᴏɴʟɪɢʜᴛDᴏʟʟs$\large{♡}$