Question

A golf ball is hit an angle of 35 degrees above the horizontal and it lands in a hole 120 m away 4.2 s later. Air resistance is negligible.?

a) What was the initial speed of the golf ball? Include a labelled diagram.
b) Determine the maximum height the ball reached. (*If you were not able to answer part (a), then assume the ball's initial speed is 25 m/s) (3 marks)

Answer 1

a) 35m/s
b) 22m

Explanation:

a) In order to determine the initial velocity of the golf ball I found the x and y components.

Since we know that it travelled 120m in 4.2s we can use this to calculate initial x velocity

initial Vx = `(120m)/(4.2s)` = 28.571m/s.

To find the initial y velocity we can use the formula

`d=Vi(t)+1/2at^2`

We know that the y displacement = 0 after 4.2s so we can plug in 0 for d and 4.2s for t.

`0 = Vi(4.2)+1/2(-9.8)(4.2^2)`

Initial Vy = 20.58

Since we now have the x and y components we can use `a^2+b^2=c^2` to find the initial velocity.

`20.58^2+28.571^2 = Vi`
`Vi = 35.211 =` 35m/s

b) To find the maximum height reached by the golf ball we can use the formula

`Vf^2 = Vi^2 + 2ad`

Since we know that the ball won't have any y velocity at its' maximum height we can substitute 0 for Vf and 20.58 for Vi.

`0^2 = 20.58^2+2(-9.8)d`
`d = 21.609 = 22m`