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Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

1
2
= 1 = sin θ k + 2 = sinθ =
k + 2

2
adjacent hypotenuse − opposite 2 k + 2 − 1 k + 1
Therefore, we have cosθ = = = =
hypotenuse hypotenuse k + 2 k + 2

Substituting the value of sin θ and cos θ in the equation for range we get:

2 2
v 2 v k + 1 1 1

R = g cos θ ∗ sinθ + k + sin θ = g ∗ + K + K + 2
k + 2 k + 2

v 2 k + 1 1 k + 1 2 v 2 k + 1 1 k + 1
= ∗ + = ∗ +
g k + 2 k + 2 K + 2 g k + 2 k + 2 K + 2

v 2 k + 1 ∗ (k + 2) v 2
= = k + 1
g k + 2 2 g

2gh
Substituting k = in the above expression, we get the maximum range of the projectile to be:
v 2

v 2 2hg
R = + 1
g v 2

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
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Copyright Diary No – 9119/2018-CO/L