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

Julia would be observing a. static electricity was caused by her sweater rubbing against her shirt and transferring electrons. she didn't touch anything metal, so that would cancel out b.she most likely didn't rub her sweater against her hair either, so that cancels out c.she also didn't scoot across the carpet while trying to take her sweater off, that cancels out d.so that leaves a.a. is correct.