Question

A bullet is fired with a velocity of 100m/s from the ground at an angle of 60° with the horizontal. calculate the horizontal range covered by the bullet. Also calculate the maximum height attained.​

Answer 1

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

$\red{\boxed{\sf Horizontal\:Range(R)=500\sqrt{3}}}$

$\red{\boxed{\sf Maximum\:Height(H_{max})=125m}}$

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

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

• Velocity of Bullet (v)=100m/s

• Angle ($\theta$)=60°

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

• Horizontal Range (R) = ?

• Maximum Height attained ($\sf{H_{max}}$)=?

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1. Horizontal Range (R) :

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

$\large\implies{\sf \frac{(100)^2×sin60°}{10}}$

[ We know sin60° = $\sf{\frac{\sqrt{3}}{2}}$ ]

$\large\implies{\sf \frac{(100)^2×\sqrt{3}}{10×2}}$

$\large\implies{\sf R=500\sqrt{3} }$

$\huge\red{♡}$$\red{\boxed{\sf Horizontal\:Range(R)=500\sqrt{3}}}$

_______________________________________

$\Large\underline{\sf 2.Maximum\: Height\:(H_{max})}$

$\large{♡}$$\large{\boxed{\sf H_{max}=\frac{u^2sin^2\theta}{2g} }}$

$\large\implies{\sf \frac{(100)^2×sin^230°}{2g}}$

[ We know sin30° =1/2 ]

$\large\implies{\sf \frac{(100)^2×1×1}{2×10×2×2}}$

$\large\implies{\sf H_{max}=125m}$

$\huge\red{♡}$$\red{\boxed{\sf Maximum\:Height(H_{max})=125m}}$

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Initial velocity (u) = 100 m/s.
• Angle of projection ( ${ \theta}$ ) = 60°.
• g = 10 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

RANGE OF THE PROJECTILE,

1.Applying the Range formula:-

$\large{\star \: {\boxed{ \tt R = \dfrac{u^2 \sin 2 \theta}{g}}}}$

Substituting the values,

$\large{ R = \dfrac{(100)^2 \sin 2 \times 60}{10}}$

$\large{ R = \dfrac{(100)^2 \sin 120}{10}}$

As we know,

sin 120° = √3/2.

Substituting,

$\large{R = \dfrac{10000 \times \dfrac{ \sqrt{3}}{2}}{10}}$

$\large{R = \dfrac{ \cancel{10000} \times \dfrac{ \sqrt{3}}{2}}{ \cancel{10}}}$

$\large{R = 1000 \times \dfrac{ \sqrt{3}}{2}}$

$\large{R = \cancel{1000} \times \dfrac{ \sqrt{3}}{ \cancel{2}}}$

The result comes as,

$\huge{\boxed{\boxed{ \tt R = 500 \sqrt{3} \: meters}}}$

$\rule{300}{1.5}$

MAXIMUM HEIGHT OF PROJECTILE

2.Applying Maximum height formula,

$\large{\star \: {\boxed{ \tt H = \dfrac{u^2 \sin^2 \theta}{2g}}}}$

Substituting the values,

$\large{ H = \dfrac{(100)^2 \sin^2 60}{2 \times 10}}$

As sin 60 = √3/2

Squaring,

sin² 60 = (√3/2)².

sin² 60 = 3/4.

Substituting it in the above equation,

$\large{H = \dfrac{10000 \times \dfrac{3}{4}}{20}}$

$\large{H = \dfrac{10000 \times 3}{80}}$

$\large{H = \dfrac{ \cancel{10000} \times 3}{ \cancel{80}}}$

$\large{H = 125 \times 3}$

$\huge{\boxed{\boxed{ \tt H = 375 \: meters}}}$

So,the Range of projectile is 500√3 meters, and,

Height attained by the projectile is 375 meters.