We can use relation: $v ^{2} = v _{0}^{2} + 2ah$ And I assume a = 9.8 m/s^2 1. Velocity before contact: Initial velocity is zero anyway. So, final velocity before contact is: $\sqrt{2ah}$ Vi = 8.85 m/s 2. Velocity after contact: Here, final velocity after rebound is zero - at the peak. So, the relation is same as above Vf = -6.26 m/s (-ve since it is going up) a. Average acceleration during contact: a = (vf - vi)/t = (-6.2 -8.8)/0.012 = -1259.61 m/s^2 b. Clearly, the acceleration is up. This is shown by the -ve sign (since i took upwards as -ve). In addition, the ball's velocity was changed from downward to upward direction. This can happen when acc is upward.