A ball is projected at a speed of 40 m/s at an angle of 60^ ^ ^ to the horizontal. Determine the range of the ball. Take g = 10 m/s^2 .

Correct answer: C. 80 3 m The horizontal range of a projectile is given by the formula: R = (u^2 (2θ))/(g) Substituting the given values u = 40 m/s , θ = 60^ ^ ^ , and g = 10 m/s^2 : R = (40)^2 (2 × 60^ ^ ^ ) 10 R = 1600 (120^ ^ ^ ) 10 Since (120^ ^ ^ ) = 3 2 : R = 160 × 3 2 = 80 3 m

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Rest and Motion Kinematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rest and Motion Kinematics

A ball is projected at a speed of 40 m/s at an angle of 60^ ^ ^ to...

A ball is projected at a speed of

40 \text{ m/s}

at an angle of

60^{}^{{}^{\circ}}

to the horizontal. Determine the range of the ball. Take

g = 10 \text{ m/s}^2

Options

A $160\sqrt{3} \text{ m}$
B $160 \text{ m}$
C $80\sqrt{3} \text{ m}$
D $120 \text{ m}$

Correct answer

C. $80\sqrt{3} \text{ m}$

Step-by-step solution

The horizontal range of a projectile is given by the formula:

R = \dfrac{u^2 \sin(2\theta)}{g}

Substituting the given values

u = 40 \text{ m/s}

\theta = 60^{}^{{}^{\circ}}

, and

g = 10 \text{ m/s}^2

R = \dfrac{(40)^2 \sin(2 \times 60^{}^{{}^{\circ}})}{10}

R = \dfrac{1600 \sin(120^{}^{{}^{\circ}})}{10}

Since

\sin(120^{}^{{}^{\circ}}) = \dfrac{\sqrt{3}}{2}

R = 160 \times \dfrac{\sqrt{3}}{2} = 80\sqrt{3} \text{ m}

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