Answer on Question #46556 – Physics - Mechanics | Kinematics | Dynamics
Find the range of a ball which was when projected with initial velocity 29.4 m/s 29.4 \, \text{m/s} 29.4 m/s 4.9 m 4.9 \, \text{m} 4.9 m
Solution:
u = 29.4 m s − initial speed of the ball ; u = 29.4 \, \frac{\text{m}}{\text{s}} - \text{initial speed of the ball}; u = 29.4 s m − initial speed of the ball ; h = 4.9 m − maximum height that ball riches h = 4.9 \, \text{m} - \text{maximum height that ball riches} h = 4.9 m − maximum height that ball riches L − range of a ball ; L - \text{range of a ball}; L − range of a ball ;
Equation of the motion for body, thrown at angle α \alpha α
u x = u cos α ; u y = u sin α ; u_x = u \cos \alpha; \quad u_y = u \sin \alpha; u x = u cos α ; u y = u sin α ; x : L = u t cos α x: L = ut \cos \alpha x : L = u t cos α y : 0 = u t sin α − g t 2 2 y: 0 = ut \sin \alpha - \frac{gt^2}{2} y : 0 = u t sin α − 2 g t 2 u sin α = g t 2 u \sin \alpha = \frac{gt}{2} u sin α = 2 g t t = 2 u sin α g t = \frac{2u \sin \alpha}{g} t = g 2 u sin α
(2) in (1):
L = u 2 u sin α g cos α = 2 u 2 sin α cos α g L = u \frac{2u \sin \alpha}{g} \cos \alpha = \frac{2u^2 \sin \alpha \cos \alpha}{g} L = u g 2 u sin α cos α = g 2 u 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 = u sin α g t _ {1} = \frac {t}{2} = \frac {u \sin \alpha}{g} t 1 = 2 t = g u sin α
y: (half of the flight): h 1 = u t 1 sin α − g t 1 2 2 h_1 = ut_1 \sin \alpha - \frac{gt_1^2}{2} h 1 = u t 1 sin α − 2 g t 1 2
h = u t 1 sin α − g t 1 2 2 h = u t _ {1} \sin \alpha - \frac {g t _ {1} ^ {2}}{2} h = u t 1 sin α − 2 g t 1 2 h = u V sin α g sin α − g ( u sin α g ) 2 2 = u 2 sin 2 α 2 g h = u \frac {V \sin \alpha}{g} \sin \alpha - \frac {g \left(\frac {u \sin \alpha}{g}\right) ^ {2}}{2} = \frac {u ^ {2} \sin^ {2} \alpha}{2 g} h = u g V sin α sin α − 2 g ( g u s i n α ) 2 = 2 g u 2 sin 2 α
from (4)
sin 2 α = 2 g h u 2 sin α = 2 g h u 2 \sin^ {2} \alpha = \frac {2 g h}{u ^ {2}} \quad \sin \alpha = \sqrt {\frac {2 g h}{u ^ {2}}} sin 2 α = u 2 2 g h sin α = u 2 2 g h
Pythagorean Identity
sin 2 α + cos 2 α = 1 \sin^ {2} \alpha + \cos^ {2} \alpha = 1 sin 2 α + cos 2 α = 1 cos 2 α = 1 − sin 2 α \cos^ {2} \alpha = 1 - \sin^ {2} \alpha cos 2 α = 1 − sin 2 α
(5)in(6):
cos 2 α = 1 − 2 g h u 2 cos α = 1 − 2 g h u 2 \cos^ {2} \alpha = 1 - \frac {2 g h}{u ^ {2}} \quad \cos \alpha = \sqrt {1 - \frac {2 g h}{u ^ {2}}} cos 2 α = 1 − u 2 2 g h cos α = 1 − u 2 2 g h
(5)and(7)in(3):
L = 2 u 2 sin α cos α g = 2 u 2 ( 2 g h u 2 ) ( 1 − 2 g h u 2 ) g = 2 ( 2 g h ) ( u 2 − 2 g h ) g = 2 ( 2 ⋅ 9.8 m s 2 ⋅ 4.9 m ) ( ( 29.4 m s ) 2 − 2 ⋅ 9.8 m s 2 ⋅ 4.9 m ) 9.8 m s 2 = 55 m L = \frac {2 u ^ {2} \sin \alpha \cos \alpha}{g} = \frac {2 u ^ {2} \sqrt {\left(\frac {2 g h}{u ^ {2}}\right) \left(1 - \frac {2 g h}{u ^ {2}}\right)}}{g} = \frac {2 \sqrt {(2 g h) (u ^ {2} - 2 g h)}}{g} = \frac {2 \sqrt {\left(2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot 4 . 9 m\right) \left(\left(2 9 . 4 \frac {m}{s}\right) ^ {2} - 2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot 4 . 9 m\right)}}{9 . 8 \frac {m}{s ^ {2}}} = 5 5 m L = g 2 u 2 sin α cos α = g 2 u 2 ( u 2 2 g h ) ( 1 − u 2 2 g h ) = g 2 ( 2 g h ) ( u 2 − 2 g h ) = 9.8 s 2 m 2 ( 2 ⋅ 9.8 s 2 m ⋅ 4.9 m ) ( ( 29.4 s m ) 2 − 2 ⋅ 9.8 s 2 m ⋅ 4.9 m ) = 55 m
Answer: the horizontal range of object will be 55 m 55\mathrm{m} 55 m
#339914 on Dec 2023