(^438) „„„„„ A Textbook of Engineering Mechanics
Example 20.22. The velocity of a particle, at its greatest height is 25 / times of its velocity
at half of its greatest height. Show that the angle of projection is 60°.
Solution. Given : Velocity of the particle at its greatest height = 2/5× Velocity at half of its
greatest height.
Let u = Initial velocity of projection, and
α= Angle of projection with the horizontal.
We know that velocity of a projectile at its greatest height
= Horizontal component of velocity of projection
= u cos (^) α ...(i)
We also know that the maximum height of projection.
(^22) sin
2
u
g
α

and half of the greatest height
1sin^2222 sin
22 4
uu
h
gg
αα
=× =
∴ Velocity of projectile at half of the greatest height,
22
(^22) –2 –2 sin
4
u
vu ghu g
g
α
==×
22
(^2) – sin
2
u
u
α
= ...(ii)
Now substituting the values from equations (i) and (ii) in the given equation,
22
cos^22 – sin
52
u
uu
α
α= ×
22–sin^222
52
uu α
=×
Squaring both sides,
222 2 2
(^22) cos^22 – sin (2 – sin )
52 5
uu u
u
αα
α= × =
5 cos^2 α = 2 – sin^2 α
or 5 (1 – sin^2 α) = 2 – sin^2 α ...(sin^2 α + cos^2 α = 1)
4sin^223 or sin^3 0.75
4
α= α= =
∴ sin α = 0.866 or α = 60° Ans.
20.11. TIME OF FLIGHT OF A PROJECTILE ON AN INCLINED PLANE
Fig. 20.17. Projectile on an inclined plane.
Consider a projectile projected from O on an upward inclined plane OA. Let the projectile
strike B as shown in Fig. 20.17.