Question

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.)

Answer 1

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