The ball is thrown at a speed of 30 m s at an angle of 45 to the horizon after a while the speed of the ball will

The ball is thrown at a speed of 30 m s at an angle of 45 to the horizon after a while the speed of the ball will be directed at an angle of 30 to the horizon to find the time.

Given:
V0 = 30m / s,
α1 = 45 °,
α2 = 30 °;
Find: t -?
The horizontal component of the speed of a body thrown at an angle, without taking into account the resistance, does not change during movement:
Vx = V0x = V0 * cos (α1) = V0 * cos (45 °) = V0 / sqrt (2);
Only the vertical component of the speed changes:
Vy = V0y – g * t = V0 * sin (α1) – g * t = V0 / sqrt (2) – g * t;
Here
g = 10m / s ^ 2 – free fall acceleration;
The speed will be at an angle of 30 ° to the horizon twice – when moving up and down;
tg (α2) = Vy / V0x = (V0y – g * t1) / V0x = 1 – (g * t1) / V0x;
and
tg (-α2) = Vy / V0x = (V0y – g * t2) / V0x = 1 – (g * t2) / V0x;
From here we find:
t1 = (V0x / g) * [1 – tg (α2)];
t2 = (V0x / g) * [1 + tan (α2)];
Substitute the values:
t1 = 0.89s;
t2 = 3.35s.