(a)
θ = angle from the goal line = 52.0 deg
v₀ = initial velocity of launch = 29 m/s
consider the motion of the ball along the vertical direction
Y = vertical displacement = 0 m
a = acceleration = - 9.8 m/s²
v' = initial velocity along the Y-direction = v₀ Sinθ = (29) Sin52 = 22.85 m/s
t = hang time
using the equation
Y = v' t + (0.5) a t²
0 = (22.85) t + (0.5) (- 9.8) t²
t = 4.7 sec
b)
Consider the motion along horizontal direction
X = displacement along X-direction
t = time taken = 4.7 sec
v'' = initial velocity along the X-direction = v₀ Cosθ = (29) Cos52 = 17.85 m/s
displacement along the X-direction is given as
X = v'' t
X = (17.85) (4.7)
X = 83.89 m
