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ch4aika [34]
3 years ago
11

A golf 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.)
Physics
1 answer:
finlep [7]3 years ago
6 0

consider the motion in x-direction

v_{ox} = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

a_{x} = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = v_{ox} t + (0.5) a_{x} t²

100 =  v_{ox} (4.60)

v_{ox} = 21.7 m/s


consider the motion along y-direction

v_{oy} = initial velocity in y-direction = ?

Y = vertical displacement  = 0 m

a_{y} = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = v_{oy} t + (0.5) a_{y} t²

0 = v_{oy} (4.60) + (0.5) (- 9.8) (4.60)²

v_{oy} = 22.54 m/s

initial velocity is given as

v_{o} = sqrt((v_{ox})² + (v_{oy})²)

v_{o} = 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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