Please answer question 3 Problem 3: A baseball of mass m, = 0.36 kg is thrown at another ball hanging fromthe ceiling by a length of string L = 1.45 m. The second ball m2 = 0.98 kg is initially atrest while the baseball has an initial horizontal velocity of V, = 3.5 m/s. After thecollision the first baseball falls straight down (no horizontal velocity).Randomized VariablesLm1 = 0.36 kgm2 = 0.98 kgL = 1.45 mV1 = 3.5 m/sm,m,Part (a) Select an expression for the magnitude of the closest distance from the ceiling the second ball will reach d.SchematicChoice :(m,V)2d2m2gd = Ld = L--2m292m29d = L -2gPart (b) What is the angle that the string makes with the vertical at the highest point of travel in degrees?Numeric : A numeric value is expected and not an expression.

Here according to conservation of momentumwe havem1v1=m2v20.36×3.5=0.98×v2v2=1.28m/sAfter collision…

Problem 3: A baseball of mass m, = 0.36 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.45 m. The second ball m2 = 0.98 kg is initially at rest while the baseball has an initial horizontal velocity of V, = 3.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity). Randomized Variables m = 0.36 kg m2 = 0.98 kg L = 1.45 m V = 3.5 m/s m, m, Part (a) Select an expression for the magnitude of the closest distance from the ceiling the second ball will reach d. SchematicChoice : d = L-- 2m2g (m,V)? d = d = L- 2m29 2m2g d = L- 2g Part (b) What is the angle that the string makes with the vertical at the highest point of travel in degrees? Numeric : A numeric value is expected and not an expression. 0 =