Suppose that a shot putter can put a shot at the worldclass speed v0 = 13.00 m/s and at a height of 2.160 m. what horizontal distance would the shot travel if the launch angle θ0 is (a) 45.00° and (b) 42.00°? the answers indicate that the angle of 45°, which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

Velocity v0 = 13.00m/s
Height h = 2.16 m
At the maximum height V = 0,
a) θ = 45 degrees
Calculating the velocity while reaching the peak
Vf =  v0sinθ = 13 x sin45 = 9.19 m/s
Calculating the time taken, V = Vf + gt => Vf - gt = 0 => t = Vf / g 
=> t = 9.19 / 9.81 => t = 0.937s
time taken for the descent will be same t = 0.937s
Total Time T = 0.937 + 0.937 = 1.874s
Now the horizontal velocity Vh = v0cosθ = 13 x cos45 = 9.19 m/s
horizontal distance s = Vh x T = 9.19 x 1.874 = 17.22 meters
b) θ = 42 degrees
Calculating the velocity while reaching the peak
Vf =  v0sinθ = 13 x sin42 = 8.67 m/s
Calculating the time taken, V = Vf + gt => Vf - gt = 0 => t = Vf / g 
=> t = 8.67 / 9.81 => t = 0.883s
time taken for the descent will be same t = 0.883s
Total Time T = 0.883 + 0.883 = 1.767s
Now the horizontal velocity Vh = v0cosθ = 13 x cos42 = 9.66 m/s
horizontal distance s = Vh x T = 9.66 x 1.767 = 17.07 meters

a) s = 17.22 m
b) s = 17.07 m

He can play two games. 13/4=3.253.25*2=6.5