Question

a body is projected with a velocity 60m/s at 30° to horizontal then it's range is
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Answer 1

• Velocity with which a body is projected , u = 60 m/s.

• Angle , θ = 30°

• g is the acceleration due to gravity, g = 10 m/s²

$range ,\: R = \dfrac{ {u}^{2} \sin(2 \theta ) }{g} \\ \\ R = \dfrac{ {(60) }^{2} \times \sin(2 \times 30) }{10} \\ \\ R = \dfrac{3600 \times \sin(60) }{10} \\ \\ R = \dfrac{3600 \times \sqrt{3} }{10 \times 2} \\ \\ R = 360 \times \dfrac{ \sqrt{3} }{2} \\ \\ R = 180 \times \sqrt{3} \: m$

Range is 180×√3 m.

Answer 2

Given :-

• Initial velocity = 60 m/s
• Angle of projection = 30°
• Acceleration due to gravity = 9.8 m/s²

To find :-

• Horizontal Range

Solution :-

$\sf Horizontal\: range = \dfrac{u^2 sin2\theta}{g}$

• u = 60 m/s
• θ = 30°
• g = 9.8 m/s²

Substituting the value in formula :-

→ $\sf R = \dfrac{u^2 sin2\theta}{g}$

→ $\sf R = \dfrac{60^2 \times sin2(30)}{9.8}$

→ $\sf R = \dfrac{3600 \times sin 60 }{9.8}$

→ $\sf R = \dfrac{3600 \times \frac{\sqrt{3}}{2} }{9.8}$

→ $\sf R = \dfrac{3117.69}{9.8}$

→ $\sf R = 318.13$

Horizontal range = 318.13 m