Chapter 20 : Projectiles „„„„„ 427

(^22) 2(1cot ) 2 1^1 tan 2
4
ugh=+α= + θgh⎛⎞⎜⎟
⎝⎠
... Substituting cot^221 tan
4
⎛⎞
⎜⎟α= θ
⎝⎠
2tan^12
2
gh
⎛⎞
=+ θ⎜⎟
⎝⎠
Ans.
Example 20.11. A projectile is aimed at a mark on the horizontal plane through the point
of projection. It falls 12 metres short when the angle of projection is 15° ; while it overshoots the
mark by 24 metres when the same angle is 45°. Find the angle of projection to hit the mark. Assume
no air resistance.
Solution. Given : When angle of projection with the horizontal (α 1 ) = 15°, horizontal
range (R 1 ) = R – 12 m and when angle of projection with the horizontal (α 2 ) = 45°, horizontal range
(R 2 ) = R + 24 m (where R is the horizontal range).
Let u= Velocity of projection, and
α= Angle of projection to hit the mark.
Fig. 20.7.
We know that horizontal range of the projectile when α= ° 15 ,
(^22)
1
1
u sin 2 u sin (2 15 )
R
gg
α ×°

∴
(^22) sin 30 0.5
(–12)
uu
R
gg
°×
== ...(i)
Similarly
2 2
2
2
u sin 2 u sin (2 45 )
R
gg
α ×°

(^22) sin 90 1
(24)
uu
R
gg
°×
+= = ...(ii)
Dividing equation (i) by (ii),
–12 0.5
24 1
R
R

• 
  

  or R – 12 = 0.5R + 12
  ∴ 0.5 R= 12 + 12 = 24 or
  24
  48 m
  0.5
  R==
  Substituting the value of R in equation (i),
  (^22) 0.5
  48 – 12
  2
  uu
  gg
  ×

  
  

  ∴ u^2 = 36 × 2g = 72g