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

Correct answer: C. 60 The maximum height H reached by a projectile is given by the formula: H = (u^2 ^2 θ)/(2g) Given u = 40 m/s , θ = 60^ ^ ^ , and g = 10 m/s^2 . Substituting the values: H = (40)^2 ( 60^ ^ ^ )^2 2 × 10 H = 1600 × ( 3 2 )^2 20 H = (1600 × (3)/(4))/(20) H = (1200)/(20) = 60 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 maximum height reached by the ball in m. Take

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

Options

A $138.5$
B $80$
C $60$
D $20$

Correct answer

C. $60$

Step-by-step solution

The maximum height

H

reached by a projectile is given by the formula:

H = \dfrac{u^2 \sin^2 \theta}{2g}

Given

u = 40 \text{ m/s}

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

, and

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

.

Substituting the values:

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

H = \dfrac{1600 \times \left(\dfrac{\sqrt{3}}{2}\right)^2}{20}

H = \dfrac{1600 \times \dfrac{3}{4}}{20}

H = \dfrac{1200}{20} = 60 \text{ m}

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