Motion In A Plane

d) Horizontal Range : -
The horizontal distance travelled by the projectile while it touches the point on the same level of the point of projection is called horizontal range.
    In the horizontal direction, velocity remains constant.
          \( \therefore Velocity = \frac{{Displacement}} {{Time}} \) 
           \( u\cos \alpha = \frac{R} {T} \) 
          \( R = u\cos \alpha .T \) 
          \( R = \left( {u\cos \alpha } \right)\left( {\frac{{2u\sin \alpha }} {g}} \right) \)  (or)  \( R = \frac{{u^2 \sin 2\alpha }} {g} \)   

e) Angle of Projection for the Maximum Range of the Projectile
    Range, R = \( \frac{{u^2 \sin \,\,2\,\alpha }} {g} \) 
    The range R for a given velocity u is maximum
    if sin 2\(\alpha\) = 1     (or)    2\(\alpha\) =\( \frac{\pi } {2} \)  (or)     \(\alpha\)= \( \frac{\pi } {2} \) = 45^o
    Corresponding to this value of a, the maximum range is  R_maximum =\( \frac{{u^2 }} {g} \) ........... (4.7)

f) Relation between range and maximum height 
    R = \( \frac{{v^2 \sin 2\theta }} {g} \)    H = \( \frac{{v^2 \sin 2\theta }} {{2g}} \)
        \( \frac{H} {R} = \frac{{\left[ {\frac{{v^2 \sin ^2 \theta }} {{2g}}} \right]}} {{\left[ {\frac{{v^2 \sin 2\theta }} {g}} \right]}} = \frac{1} {4}\tan \theta \) (or)  \( \boxed{Tan\theta = \frac{{4H}} {R}} \)      
Sub case : Condition for range of projectile to be equal to maximum height attained by it
    Tan \(\theta\) = 4,  \( \tan \alpha = 4 \) ,  \( \alpha = \tan ^{ - 1} 4 = 76^0 \)    

g) There are two angles of projection for same range  
    Replacing by (90⁰ –\(\theta\)) in the formula of range we get
          \( R = \frac{{u^2 \sin ^2 (90^0 - \theta )}} {g} = \frac{{u^2 \sin \left( {180 - 2\theta } \right)}} {g} = \frac{{u^2 \sin 2\theta }} {g} = R \)
Thus, for a given velocity of projection, a projectile has the same range for angle of projection \(\theta\) and \((90^0-\theta)\)
Note : In the above case range of two projections is same but time of flights are different.

h) Relation between times of flights and range in case of projectiles having same range.
         

    Time of flight for angle of projection\(\theta\).
   \( T_1 = \frac{{2u\sin \theta }} {g} \) and time of the flight for angleof projetion \( (90^0 - \theta ) \),
    T₂ = \( \frac{{2u\sin \left( {90^0 - \theta } \right)}} {g} = \frac{{2u\cos \theta }} {g} \)
    Multiplying T₁  and T₂ we get
    \( T_1 T_2 = \frac{{2u\sin \theta }} {g} \times \frac{{2u\cos \theta }} {g} \)
    or   \( T_1 T_2 = \frac{2} {g}\left( {\frac{{u^2 \sin 2\theta }} {g}} \right) \) (or)  \( T_1 T_2 = \frac{{2R}} {g} \)    \( \frac{{T_1 }} {{T_2 }} = Tan\theta \)  

(i) If t₁ is the time taken by projectile to reach a point P at height h and t₂ is the time taken from point P to  ground level, then
       \( t_1 + t_2 = T = \frac{{2u\sin \theta }} {g} \) or  \( u\sin \theta = \frac{{g\left( {t_1 + t_2 } \right)}} {2} \)   
            

The height of point P,  \( h = u\sin \theta t_1 - \frac{1} {2}gt_1 ^2 = \frac{{g(t_1 + t_2 )}} {2}t_1 - \frac{1} {2}gt_1 ^2 \)    or  \( h = \frac{1} {2}gt_1 t_2 \)