Answer:
1) 13.377 m/s
2) 12.044 m/s
3) 8.893 m
4) 32.85 m
5) 19.05 m/s
6) 3.25 m
Explanation:
1)
V_o,x = V_o * cos (Q)
V_o,x = 18 * cos (42)
V_o,x = 13.377 m/s
2)
V_o,y = V_o * sin (Q)
V_o,y = 18 * sin (42)
V_o,y = 12.044 m/s
3)
Maximum height is reached when V,y = 0
V,y = V_o,y + a*t
0 = 12.044 - 9.81t
Solve above equation for t:
t = 1.228 s
Compute S_y @t = 1.228 s
S_y = S_o,y + V_o,y*t + 0.5*a*t^2
S_y = 1.5 + 12.044*1.228 - 4.905*1.228^2
S_y = 8.893 m
4)
Time taken for the ball to complete path:
S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 1.5
V_o,y*t + 0.5*a*t^2 = 0
12.044*t - 4.905*t^2 = 0
t = 0, t = 2.455 s
Total distance traveled in horizontal direction S_x @ t = 2.455 s
S_x = S_o,x + V_o,x*t
S_x = 0 + 13.377*2.455 = 32.85 m
5)
S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 10
10 = 1.5 + V_o,y*t -4.905*t^2 .... Eq 1
Maximum height is reached when V,y = 0
V,y = V_o,y + a*t
0 = V_o,y - 9.81t .... Eq2
Solve Eq 1 and Eq 2 simultaneously
V_o,y = 9.81*t
10 = 1.5 + 9.81*t^2 -4.905*t^2
8.5 = 4.905*t^2
t = 1.316 s
V_o,y = 12.914 m/s
Compute Velocity
V = sqrt (V_o,x^2 + V_o,y^2)
V = sqrt (14^2 + 12.914^2)
V = 19.05 m/s
6)
Total distance traveled in horizontal direction between players is 32.85m
S_x = S_o,x + V_o,x*t
S_x = 0 + 14*t = 32.85 m
t = 2.3464 s
Compute Sy @ t = 2.3464 s
S_y = S_o,y + V_o,y*t + 0.5*a*t^2
S_y = 10 - 4.905*(1.1732)^2
S_y = 3.25 m
