The ball is thrown at an angle to the horizon with an initial speed of 10 m / s. What is the maximum height of the ball?

According to the law of conservation of energy:

Eк1 + Eп1 = Eк2 + Eп2, where Eк1 and Eп1 are the energy of the ball during the throw; Eк2 and Ep2 – ball energy at maximum height.

At the moment of throwing h = 0 m, therefore Ep1 = 0.

At the maximum height, V = 0, so Ek2 = 0.

Ek1 = En2.

Eк1 = m * V0² / 2, m is the mass of the ball, V0 is the initial velocity of the ball (V = 10 m / s).

Ep2 = m * g * h, g – gravitational acceleration (g = 10 m / s²), h – maximum height.

m * V0² / 2 = m * g * h.

h = V0² / 2g = 102 / (2 * 10) = 100/20 = 5 m.