Question

Найти угол α, под которым атлету следует выпустить ядро, чтобы дальность полета ядра была максимальной. Рост атлета не учитывать. Сопротивлением воздуха пренебречь.

Answer 1

Дальность полета x=V0x*t   V0x=V0*cosa   t=2*V0*sina/g
x=v0*cosa*2*V0*sina/g так как 2*sina*cosa=sin2a то:
x=V0*sin2*a/g  максимальна дальность будет при sin2a=1. что соответствует углу a=45 градусов.

Answer 2

Finding the Angle for Maximum Projectile Range

To find the angle at which an athlete should release a projectile to achieve maximum range, we can use the principles of projectile motion. In this case, we are assuming that the athlete is throwing a projectile without considering their height and neglecting air resistance.

The maximum range of a projectile occurs when it is launched at an angle of 45 degrees with respect to the horizontal. This is a result of the symmetrical nature of projectile motion, where the time of flight and horizontal distance covered are maximized at this angle.

Therefore, the angle α at which the athlete should release the projectile to maximize its range is 45 degrees.

Please note that this answer assumes ideal conditions and neglects factors such as the athlete's height and air resistance.