a body is projected at two complementary angles with same speed if maximum height reached by it are 40 m 40\mathrm{m} 40 m 90 m 90\mathrm{m} 90 m
Solution:
We can write the equations of motion for each of the bodies and use fact that:
α = 9 0 o − β \alpha = 90^{o} - \beta α = 9 0 o − β
1.
Equations for body, thrown at angle α \alpha α
V x = V cos α ; V y = V sin α ; V _ {x} = V \cos \alpha ; V _ {y} = V \sin \alpha ; V x = V cos α ; V y = V sin α ; x : L 1 = V t cos α ( 1 ) , t − time of the flight x \colon L _ {1} = V t \cos \alpha (1), t - \text {time of the flight} x : L 1 = V t cos α ( 1 ) , t − time of the flight y : 0 = V t sin α − g t 2 2 y \colon 0 = V t \sin \alpha - \frac {g t ^ {2}}{2} y : 0 = V t sin α − 2 g t 2 V sin α = g t 2 V \sin \alpha = \frac {g t}{2} V sin α = 2 g t t = 2 V sin α g ( 2 ) t = \frac {2 V \sin \alpha}{g} (2) t = g 2 V sin α ( 2 ) ( 2 ) in ( 1 ) : L 1 = V 2 V sin α g cos α = 2 V 2 sin α cos α g (2) \text {in} (1): L _ {1} = V \frac {2 V \sin \alpha}{g} \cos \alpha = \frac {2 V ^ {2} \sin \alpha \cos \alpha}{g} ( 2 ) in ( 1 ) : L 1 = V g 2 V sin α cos α = g 2 V 2 sin α cos α
Maximum height: the time taken to reach the maximum height is equal to half of the time of flight:
t 1 = t 2 = V sin α g t _ {1} = \frac {t}{2} = \frac {V \sin \alpha}{g} t 1 = 2 t = g V sin α y : ( h a l f o f t h e f l i g h t ) : h 1 = V t 1 sin α − g t 1 2 2 y: (half \ of \ the \ flight): h_1 = Vt_1 \sin \alpha - \frac{gt_1^2}{2} y : ( ha l f o f t h e f l i g h t ) : h 1 = V t 1 sin α − 2 g t 1 2 h 1 = V t 1 sin α − g t 1 2 2 h_1 = Vt_1 \sin \alpha - \frac{gt_1^2}{2} h 1 = V t 1 sin α − 2 g t 1 2 h 1 = V V sin α g sin α − g ( V sin α g ) 2 2 = V 2 sin 2 α 2 g h_1 = V \frac{V \sin \alpha}{g} \sin \alpha - \frac{g \left(\frac{V \sin \alpha}{g}\right)^2}{2} = \frac{V^2 \sin^2 \alpha}{2g} h 1 = V g V sin α sin α − 2 g ( g V s i n α ) 2 = 2 g V 2 sin 2 α
Equations for body, thrown at angle β = 90 ∘ − α \beta = 90{}^\circ - \alpha β = 90 ∘ − α
We will use the properties of trigonometric:
sin ( 90 ∘ − α ) = cos α ; cos ( 90 ∘ − α ) = sin α \sin(90{}^\circ - \alpha) = \cos \alpha; \cos(90{}^\circ - \alpha) = \sin \alpha sin ( 90 ∘ − α ) = cos α ; cos ( 90 ∘ − α ) = sin α
2.
Equations for body, thrown at angle α \alpha α L 1 L_1 L 1
V x = V cos β = V sin α ; V y = V sin β = V cos α ; V_x = V \cos \beta = V \sin \alpha; \quad V_y = V \sin \beta = V \cos \alpha; V x = V cos β = V sin α ; V y = V sin β = V cos α ; x : L 1 = V t ′ sin α ( 1 ) ′ , t ′ − time of the flight x: L_1 = Vt' \sin \alpha \quad (1)', t' - \text{time of the flight} x : L 1 = V t ′ sin α ( 1 ) ′ , t ′ − time of the flight y : 0 = V t ′ cos α − g t 2 2 y: 0 = Vt' \cos \alpha - \frac{gt^2}{2} y : 0 = V t ′ cos α − 2 g t 2 V cos α = g t ′ 2 V \cos \alpha = \frac{gt'}{2} V cos α = 2 g t ′ t ′ = 2 V cos α g ( 2 ) ′ t' = \frac{2V \cos \alpha}{g} \quad (2)' t ′ = g 2 V cos α ( 2 ) ′ ( 2 ) in ( 1 ) : L 2 = V 2 V sin α g cos α = 2 V 2 sin α cos α g = L 1 (2) \text{in} \quad (1): L_2 = V \frac{2V \sin \alpha}{g} \cos \alpha = \frac{2V^2 \sin \alpha \cos \alpha}{g} = L_1 ( 2 ) in ( 1 ) : L 2 = V g 2 V sin α cos α = g 2 V 2 sin α cos α = L 1
It means that the maximum range in both cases are equal.
Maximum height: the time taken to reach the maximum height is equal to half of the time of flight:
t 2 = t ′ 2 = V cos α g t_2 = \frac{t'}{2} = \frac{V \cos \alpha}{g} t 2 = 2 t ′ = g V cos α y : ( h a l f o f t h e f l i g h t ) : h 2 = V t 2 cos α − g t 2 2 2 y: (half \ of \ the \ flight): h_2 = Vt_2 \cos \alpha - \frac{gt_2^2}{2} y : ( ha l f o f t h e f l i g h t ) : h 2 = V t 2 cos α − 2 g t 2 2 h 2 = V t 2 cos α − g t 2 2 2 h_2 = Vt_2 \cos \alpha - \frac{gt_2^2}{2} h 2 = V t 2 cos α − 2 g t 2 2 h 2 = V V cos α g cos α − g ( V cos α g ) 2 2 = V 2 cos 2 α 2 g h_2 = V \frac{V \cos \alpha}{g} \cos \alpha - \frac{g \left(\frac{V \cos \alpha}{g}\right)^2}{2} = \frac{V^2 \cos^2 \alpha}{2g} h 2 = V g V cos α cos α − 2 g ( g V c o s α ) 2 = 2 g V 2 cos 2 α
So, we have a system of equations:
{ L = 2 V 2 sin α cos α g ( 3 ) h 1 = V 2 sin 2 α 2 g ( 4 ) h 2 = V 2 cos 2 α 2 g ( 5 ) \left\{ \begin{array}{l} L = \frac {2 V ^ {2} \sin \alpha \cos \alpha}{g} \quad (3) \\ h _ {1} = \frac {V ^ {2} \sin^ {2} \alpha}{2 g} \quad (4) \\ h _ {2} = \frac {V ^ {2} \cos^ {2} \alpha}{2 g} \quad (5) \end{array} \right. ⎩ ⎨ ⎧ L = g 2 V 2 s i n α c o s α ( 3 ) h 1 = 2 g V 2 s i n 2 α ( 4 ) h 2 = 2 g V 2 c o s 2 α ( 5 ) ( 4 ) ( 5 ) ; h 1 h 2 = V 2 sin 2 α 2 g ∗ 2 g V 2 cos 2 α = sin 2 α cos 2 α = 40 m 90 m = 4 9 \frac {(4)}{(5)}; \frac {h _ {1}}{h _ {2}} = \frac {V ^ {2} \sin^ {2} \alpha}{2 g} * \frac {2 g}{V ^ {2} \cos^ {2} \alpha} = \frac {\sin^ {2} \alpha}{\cos^ {2} \alpha} = \frac {4 0 m}{9 0 m} = \frac {4}{9} ( 5 ) ( 4 ) ; h 2 h 1 = 2 g V 2 sin 2 α ∗ V 2 cos 2 α 2 g = cos 2 α sin 2 α = 90 m 40 m = 9 4
apply the square root of both sides of the equation:
sin 2 α cos 2 α = 4 9 = > sin α cos α = 2 3 = tan α ; \frac {\sin^ {2} \alpha}{\cos^ {2} \alpha} = \frac {4}{9} = > \frac {\sin \alpha}{\cos \alpha} = \frac {2}{3} = \tan \alpha ; cos 2 α sin 2 α = 9 4 => cos α sin α = 3 2 = tan α ; ( 3 ) ( 4 ) ; L h 1 = 2 V 2 sin α cos α g ∗ 2 g V 2 sin 2 α \frac {(3)}{(4)}; \frac {L}{h _ {1}} = \frac {2 V ^ {2} \sin \alpha \cos \alpha}{g} * \frac {2 g}{V ^ {2} \sin^ {2} \alpha} ( 4 ) ( 3 ) ; h 1 L = g 2 V 2 sin α cos α ∗ V 2 sin 2 α 2 g L h 1 = 2 cos α 1 ∗ 2 sin α = 4 tan α ; \frac {L}{h _ {1}} = \frac {2 \cos \alpha}{1} * \frac {2}{\sin \alpha} = \frac {4}{\tan \alpha}; h 1 L = 1 2 cos α ∗ sin α 2 = tan α 4 ; 2 3 = tan α ; = > \frac {2}{3} = \tan \alpha ; = > 3 2 = tan α ; => L = 4 h 1 tan α = 3 ∗ 4 ∗ 40 m 2 = 240 m L = \frac {4 h _ {1}}{\tan \alpha} = \frac {3 * 4 * 4 0 m}{2} = 2 4 0 m L = tan α 4 h 1 = 2 3 ∗ 4 ∗ 40 m = 240 m
Answer: maximum range L = 240 m L = 240m L = 240 m
