Question

A golfer is teeing off on a 170.0 m long par 3 hole. The ball leaves with a velocity of 40.0 m/s at50.0° to the horizontal. Assuming that she hits the ball on a direct path to the hole, how far fromthe hole will the ball land (no bounces or rolls)?

Answer 1

Given data:

Initial velocity of the ball;

`u=40.0\text{ m/s}`

Launch angle;

`\theta=50.0\degree`

Distance between the launch site and the hole;

`D=170.0\text{ m}`

The range of the projectile is given as,

`R=(u^2\sin 2\theta)/(g)`

Here, g is the acceleration due to gravity.

Substituting all known values,

`\begin{gathered} R=\frac{(40\text{ m/s})^2*\sin (2*50\degree)}{9.8\text{ m/s}^2} \\ \approx160.78\text{ m} \end{gathered}`

The distance between the hole and landing site is given as,

`d=D-R`

Substituting all known values,

`\begin{gathered} d=(170\text{ m})-(160.78\text{ m}) \\ =9.22\text{ m} \end{gathered}`

Therefore, the ball will land 9.22 m away from the hole.