Question

a body of mass 2kg is projected at an angle of 45degree with the horizontal. if the maximum height is 50m then its horizontal range is? ​

Answer 1

Answer :

• Horizontal range of the body, R = 200 m.

Explanation :

Given :

• Angle of projectile, θ = 45°
• Maximum height of the body, h₍ₘₐₓ₎ = 50 m
• Acceleration due to gravity, g = 10 m/s²

To find :

• Horizontal range of the body, R = ?

Knowledge required :

• Formula for Maximum height reached by a particle, h₍ₘₐₓ₎ = (u²sin²θ)/2g
• Formula for Horizontal range of a particle, R = (u²sin2θ)/g

Where :

⠀⠀⠀⠀⠀⠀• u = Initial velocity of the particle.

⠀⠀⠀⠀⠀⠀• θ = Angle of projection.

⠀⠀⠀⠀⠀⠀• g = Acceleration due to gravity.

Concept :

In the question, we are provided with the angle of projectile and the maximum height reached by the particle. So, from the equation of maximum height, we can obtain the Initial velocity of the body and then by using the equation for horizontal range, we can find out the range of the body.

Solution :

Initial velocity of the body :

⠀⠀By using the equation for Maximum height reached by the body and substituting the values in it, we get :

⠀⠀⠀=> h₍ₘₐₓ₎ = (u²sin²θ)/2g

⠀⠀⠀=> 50 = (u²sin²45°)/2(9.8)

⠀⠀⠀=> 50 = (u²(1/√2²)/19.6⠀⠀ [∵ sin45° = 1/√2]

⠀⠀⠀=> 50 = (u²/2)/19.6

⠀⠀⠀=> u = √(39.2 × 50)

⠀⠀⠀=> u = 44.27

⠀⠀⠀⠀⠀⠀∴ u = 44.27 m/s

Hence, the Initial velocity of the body is 44.27 m/s.

Horizontal range of the body :

⠀⠀By using the equation for Horizontal range of the body and substituting the values in it, we get :

⠀⠀⠀=> R = (u²sin2θ)/g

⠀⠀⠀=> R = [(44.26)² × sin(2 × 45°)]/9.8

⠀⠀⠀=> R = (1960 × sin90°)/9.8

⠀⠀⠀=> R = (1960 × 1)/9.8⠀⠀ [∵ sin90° = 1]

⠀⠀⠀=> R = 1960/9.8

⠀⠀⠀=> R = 200

⠀⠀⠀⠀⠀⠀∴ R = 200 m

Hence, the horizontal range of the body is 200 m.

Answer 2

Given :-

a body of mass 2kg is projected at an angle of 45degree with the horizontal. if the maximum height is 50m

To Find :-

Horizontal range

Solution :-

We know that

$\sf Maximum_{Height} = \dfrac{u^2\sin^2\theta}{2\times g}$

$\sf 50=\dfrac{u^2 sin^2 45}{2 \times 9.8}$

$\sf 50=\dfrac{u^2 sin^245}{19.6}$

$\sf 50\times 19.6 = u^2\times \dfrac{1}{\sqrt{2}}$

$\sf 980 = \dfrac{u^2}{\sqrt{2}}$

$\sf u = 44.27$

Now

$\sf Horizonal = \dfrac{(44.26)^2\times sin(45 \times 2)}{9.8}$

$\sf Horizontal = \dfrac{1960 \times sin(45 \times 2)}{9.8}$

$\sf Horizontal = \dfrac{1960\times 1}{9.8}$

$\sf Horizontal = 200 \; m$