The ball was thrown at an initial speed of 9 m / s at an angle of 30 degrees to the horizon. Determine the height of the ball.

Given:
v0 = 9m / s,
α = 30 °,
Find: h -?
From the equation for uniformly accelerated motion, we write what the lift height is equal to:
h = [(v0y) ^ 2 – (vy) ^ 2] / (2 * g);
here
g = 10m / s ^ 2 – acceleration of gravity,
v0y – vertical component of speed at the beginning of movement,
vy is the vertical component of the velocity at the height h;
Since at the highest point
vy = 0,
then
h = 1 / (2 * g) * (v0y) ^ 2;
v0y = v0 * sin (α) = v0 / 2 = 4.5m / s;
h = 1 / (20) * 20.25 = 1.0125m.