Aprojectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. the two protons thenmove along perpendicular paths, with the projectile path at 60 from the original direction. after the collision, what are the speedsof (a) the target proton and (b) the projectile proton

Because the two paths are perpendicular, therefore the target proton's new path must be at 30 degrees from the original direction. 

Using the law of conservation of momentum about the original direction: 
m (400 m/s) = m (v1) cos(60) + m (v2) cos(30) 
Cancelling m since the two protons have similar mass.
(v1)cos(60) + (v2)cos(30) = 500 m/s                         ---> 1

Now by using the law conservation of momentum perpendicular to the original direction: 
m (0 m/s) = m (v1) sin(60) – m (v2) sin(30) 
Which simplifies to:
(v1)sin(60) - (v2)sin(30) = 0 m/s                                
v2 = v1 * sin(60) / sin(30) = v1 * sqrt(3)                  ---> 2

Plugging equation 2 to equation 1: 
(v1) (1/2) + (v1 * sqrt(3)) sqrt(3)/2 = 500 m/s 
(1/2) (v1) + (3/2) (v1) = 500 m/s 
2 (v1) = 500 m/s 
v1 = 250 m/s 

Thus, from equation 2:

v2 = v1*sqrt(3) = (250 m/s) sqrt(3) = 433.01 m/s 

So,
A. The target proton's speed is about 433 m/s 
B. The projectile proton's speed is 250 m/s

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s