Asoftball is thrown from the origin of an x-y coordinate system with an initial speed of 18 m/s at an angle of 35 degrees above the horizontal. part a) find the x positions of the softball at the times t=.50s, 1.0s, 1.5s, and 2.0s. part b) find the y positions of the softball at the times t=.50s, 1.0s, 1.5s, and 2.0s.

Vo = 18 m/s
angle 35 degrees

1) Components of the initial velocity
 
Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s

2) Equations of postion:

x = Vox*t
y = Voy*t - gt^2 / 2

3) Calculations

A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s

x = 14.74 * t

t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m

t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m

t = 1.5s => x = 22.11 m

t = 2s => x = 29.48 m

B)

y = Voy*t - gt^2 / 2

Voy = 10.32 m/s
g = 10 m/s (approximation)

y = 10.32*t - 5t^2

t = 0.5 s=> y = 3.91m

t = 1 s => y = 5.32m

t = 1.5 s => y = 4.23m

t = 2 s => y = 0.64 m