The athlete pushes the core at an initial speed of 18 m / s at an angle of 45 ° to the horizon.

The athlete pushes the core at an initial speed of 18 m / s at an angle of 45 ° to the horizon. determine the flight time of the nucleus.

∠α = 45 °.

g = 10 m / s2.

V0 = 18 m / s.

t -?

The motion of the nucleus can be divided into two types: horizontally, it moves uniformly with a speed Vх = V0 * cosα, vertically uniformly accelerated with gravitational acceleration g and an initial speed V0у = V0 * sinα.

The nucleus moves vertically up to a complete stop, and then begins to move down.

The flight time of the nucleus is expressed by the formula: t = t1 + t2, where t1 is the time of the ascent of the nucleus to the maximum height, t2 is the time of the descent of the nucleus from the maximum height.

t1 = t2, so t = 2 * t1.

The core ascent time t1 is expressed by the formula: t1 = (V0у – Vу) / g. Since the kernel has stopped, Vу = 0.

t1 = V0у / g = V0 * sinα / g.

t = 2 * V0 * sinα / g.

t = 2 * 18 m / s * sin45 ° / 10 m / s2 = 2.6 s.

Answer: the flight time of the nucleus will be t = 2.6 s.