Chapter 20 : Projectiles „„„„„ 437

We know that the vertical component of initial velocity

=u sin α

and vertical component of the final velocity after t seconds

=v sin θ

This change in velocity (i.e. from u sin αto v sin θ) is becasue of the retardation (g) due to

gravity.

∴ v^2 sin^2 θ = u^2 sin^2 α – 2gh ...(Q^ v^2 = u^2 – 2as) ...(i)

or vusinθ=^22 sin α– 2gh ...(ii)

We also know that the horizontal component of these two velocities does not change.

∴ v cos θ = u cos α ...(iii)

Squaring equation (iii) and adding to equation (i),

v^2 sin^2 θ + v^2 cos^2 θ =u^2 sin^2 α – 2gh + u^2 cos^2 α

v^2 (sin^2 θ + cos θ)=u^2 (sin^2 α + cos^2 α) – 2gh

or v^2 =u^2 – 2gh ...(Q sin^2 θ + cos^2 θ = 1)

∴ vu gh=^2 –2
This angle which the particle makes with the horizontal at P may be found out by dividing the
equation (ii) by (iii) i.e.,

sin^22 sin – 2

cos cos

v ugh

vu

θ α

=

θα

∴

Vertical velocity at a height

tan

Horizontal component of initial velocity

h

θ=

Example 20.21. A body is projected upwards with a velocity of 50 m/s at angle of 50° with
the horizontal. What will be its (i) velocity and (ii) direction at a height of 30 m from the point of
projection.

Solution. Given : Initial velocity of projection (u) = 50 m/s ; Angle of projection (α) = 50°
and height (h) = 30 m.

(i) Velocity of the projectile

We know that velocity of the projectile,

vu gh==××^22 –2 (50) –(2 9.8 30) = 43.7 m/s Ans.

(ii) Direction of the projectile

Let θ= Angle which the projectile makes with the horizontal.

We also know that

(^22) sin – 2 (50) sin 50 – 2 9.8 30 22
tan
cos 50 cos 50
ugh
u
α°××
θ= =
α°
2500 (0.766) – 588^2
0.9224
50 0.6428
×

×
or θ= 42.7° Ans.