How fast should you throw the ball down so that it bounces 4 m above the level from which it was thrown?

1). Let us denote the sought speed with which it is necessary to throw a ball of mass m downwards so that it bounces 4 m above the initial level h by the letter Vо. Then the kinetic energy that the ball will receive during the throw will be Wк = (m • Vо ^ 2): 2. At this height, the ball has a potential energy of interaction with the ground: Wп = m • g • h, where the constant g = 9.8 N / kg. Full mechanical energy of the ball W = Wк + Wп; W1 = (m • Vо ^ 2): 2 + m • g • h.
2). At the top flight point at an altitude of H = h + 4, which is 4 m higher than the initial level h, the ball has potential energy Wp = m • g • H and kinetic energy Wk = (m • V ^ 2): 2, where velocity V = 0 m / s, which means Wk = 0 J. The total mechanical energy of the ball will be W = Wk + Wp; W2 = m • g • (h + 4).
3). According to the law of conservation of mechanical energy W1 = W2; that is, (m • Vо ^ 2): 2 + m • g • h = m • g • (h +4); Vo ^ 2 = 8 • g; Vо ≈ 8.85 m / s. Answer: the ball must be thrown at a speed of Vо ≈ 8.85 m / s.