Question

A golf ball is hit from ground level at an angle of 25 degrees above ground level. The terrain is level. The ball lands 80.0438 meters away. What was the initial speed of the golf ball?

Answer 1

`32` m/s

Step-by-step explanation:

`\theta` = angle of launch of the ball = 25 deg

`v_(o)` = speed of launch of the golf ball = ?

Consider the motion along vertical direction

`v_(oy)` = initial velocity along y-direction =
`v_(o) Sin\theta` =
`v_(o) Sin25`

`t` = time of travel in air = ?

`Y` = vertical displacement = 0 m

`a` = acceleration due to gravity = 9.8 m/s²

Using the equation

`Y = v_(oy) t - (0.5) a t^(2)`

`0 = v_(o) Sin\theta t - (0.5) a t^(2)`

`t = (2v_(o) Sin\theta )/(a)` eq-1

Consider the motion along horizontal direction

`v_(ox)` = initial velocity along x-direction =
`v_(o) Cos\theta` =
`v_(o) Cos25`

`t` = time of travel in air = ?

`X` = horizontal displacement = 80.0438 m

Using the equation

`X = (v_(o) Cos\theta) t`

`80.0438 = (v_(o) Cos25) t`

using eq-1

`80.0438 = (v_(o) Cos25) \left ( (2v_(o) Sin25 )/(g) \right )`

`80.0438 = (v_(o) Cos25) \left ( (2v_(o) Sin25 )/(9.8) \right )`

`v_(o) = 32` m/s