Agolf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction of the initial velocity? (neglect air resistance.)

consider the motion in x-direction

= initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

= acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = t + (0.5) t²

100 =   (4.60)

= 21.7 m/s

consider the motion along y-direction

= initial velocity in y-direction = ?

Y = vertical displacement  = 0 m

= acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = t + (0.5) t²

0 = (4.60) + (0.5) (- 9.8) (4.60)²

= 22.54 m/s

initial velocity is given as

= sqrt(()² + ()²)

= sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

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