A golf ball is struck with a five iron on level ground. It lands 98.6 m away 4.59 s later. What was the direction and magnitude of the initial velocity?The magnitude of the initial velocity...

A golf ball is struck with a five iron on level ground. It lands 98.6 m away 4.59 s later. What was the direction and magnitude of the initial velocity?

The magnitude of the initial velocity is  m/s
The direction of the initial velocity is  degrees above the horizontal

 

 

 

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lfryerda | High School Teacher | (Level 2) Educator

Posted on

We know from the equations of motion that the vertical height of a ball is given by:

`h=-1/2g t^2+v_{0y}t+h_0`

where g is the acceleration due to gravity, t is the height, `v_{0y}` is the vertical component of the initial velocity of the ball and `h_0` is the initial height of the ball, which is zero.  

Now, we have that the ball lands 4.59 seconds after it was struck.  By substituting in the values for the height equation, we get:

`0=-9.8/2(4.59)^2+v_{0y}(4.59)`   solve for `v_{0y}`

`v_{0y}=9.8/2(4.59)=22.491`

With the horizontal component of the ball, we just have the uniform velocity equation, which is:

`d=v_{0x}t`

substituting in values, we get:

`98.6=v_{0x}(4.59)`  and solving to find:

`v_{0x}=98.6/4.59=21.48`

Now we can find the angle and magnitude of the initial velocity using pythagorean theorem to find:

`v_0=sqrt{v_{0x}^2+v_{0y}^2}`

`=sqrt{21.48^2+22.49^2}=31.1`

The angle above horizontal is:

`theta=tan^{-1}(v_{0y}/v_{0x})`

`=tan^{-1}(22.49/21.48)=46`  degrees

The initial velocity has magnitude of 31.1 m/s and is 46 degrees above horizontal.

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