1. The problem statement, all variables and given/known data You are standing on scales which read weight in Newtons. A 0.50 kg ball is dropped from a height of 1 m into your hands. Your hands drop from chest level to waist level during the catch, a distance of about 25 cm. Your mass is 62 kg. Assuming that you decelerate the ball uniformly during the catch, what would be the maximum reading on the scales? (Hint: The scales read 607.6 N before you caught the ball.) Weight of person = 607.6N Mass of Ball= 0.50 kg Displacement of ball when dropped = 1.0 metres Displacement of ball during deceleration = 0.25 metres 2. Relevant equations Force = Mass * acceleration Force (weight) = Mass * 9.81 Velocity (final) ^2= Velocity (initial) ^2 + 2 * acceleration * displacement Vf^2 = Vi^2 + (2 * a * d) 3. The attempt at a solution Calculate final velocity of ball after it is first dropped before it is caught: Vf= x Vi= 0 m/s a= 9.81 m/s^2 d= 1.0 metres Vf^2 = Vi^2 + (2 * a * d) x^2 = 0^2 + (2*9.81*1) x^2 = 19.62 x = 4.43 m/s Calculate acceleration (negative) of ball as it is being caught: Vf= 0 m/s Vi= 4.43 m/s a= x d= 0.25 metres Vf^2 = Vi^2 + (2 * a * d) 0^2 = 4.43^2 + (2 * x * 0.25) 0 = 19.62 + (0.50x) -0.50x = 19.62 x = 39.24 Therefore, the acceleration of the ball as it is being caught is 39.24 metres per second squared downward. Now, calculate the force of the ball: F=ma F=0.5 x 39.24 F=19.62 Now, calculate the maximum weight reading on the scale (which occurs as the ball is being caught): Weight of man + weight of ball + force of ball during deceleration 607.6 newtons + (0.5)*(9.81) + 19.62 newtons which is... 607.6 newtons + 4.905 newtons + 19.62 newtons=632.125 newtons Thus, the maximum scale reading would be 632.125 newtons. Am i right?