Question

A golfer is teeing off from the middle of a raised tee. The tee is 10.0 meters wide at its base and 3.00 meters above the 25.0 meter wide green, which is sitting on a flat mesa. Assume that the green is circular. The pin (flag) is 120 meters from the base of the tee. He hits his ball with a 7 iron, giving it a velocity of 34.0m/s at an angle of 40.0 degrees. You must show work for credit.

1) what is the vertical velocity?
2) what is the horizontal velocity?
3) what is the time of flight?
4) how high does the ball goes?
5) does he reach the green?

Answer 1

A golfer is teeing off from the middle of a raised tee. The tee is 10.0 meters wide at its base and 3.00 meters above the 25.0 meter wide green, which is sitting on a flat mesa. Assume that the green is circular. The pin (flag) is 120 meters from the base of the tee. He hits his ball with a 7 iron, giving it a velocity of 34.0 m/s at an angle of 40.0 degrees. You must show work for credit. figure attached shows the golf course.

1. The vertical component of velocity = 21.9 m/s

2.The Horizontal component of velocity = 26.1 m/s

3. The time of the flight = 4.6 secs

4. The Maximum height of the golf ball = 27.4 m

5. The golfer reached the green

Step-by-step explanation:

This is a case of projectile motion;

1. The vertical component of velocity

The vertical component of the velocity can be calculated as follows;

`v_(y)` = vsinθ

`v_(y)` is the vertical velocity

v is the initial velocity = 34 m/s

and θ is the projectile angle =
`40^(o)`

`v_(y)` = (34 m/s) x (sin
`40^(o)`)

`v_(y)` = 21.9 m/s

The vertical velocity is 21.9 m/s

2. The Horizontal component of velocity

The horizontal component of the velocity can be calculated as follows;

`v_(x)` = vcosθ

`v_(x)` is the horizontal velocity

v is the initial velocity = 34 m/s

and θ is the projectile angle =
`40^(o)`

`v_(x)` = (34 m/s) x (cos
`40^(o)`)

`v_(x)` = 26.1 m/s

The horizontal velocity is 26.1 m/s

3. The time of the flight.

The time of flight can be calculated with the relationship below;

y =
`v_(y)` t -
`(1)/(2)gt^(2)`

y is the vertical displacement (from the origin) = 0m- 3m = -3 m/s

`v_(y)` is the initial vertical velocity = 21.9 m/s

g is the acceleration due to gravity = 9.8 m/
`s^(2)`

substituting the values in the expression.

3 = (21.9 x t) - (
`(1)/(2)` x 9.8 x
`t^(2)`)

3 = 21.9 t - 4.9
`t^(2)`

rearranging the equation we have;

4.9
`t^(2)` - 21.9 t -3 = 0

using the quadratic formula to solve the equation we have;

t = 21.9 ±
`\frac{\sqrt{(-21.9) ^(2) -(4 * 4.9*(-3) )} }{2 * 4.9}`

t = −0.133 or 4.6

but time can not be negative so t = 4.6 secs.

The time of flight is 4.6 secs

4. The Maximum height of the golf ball

The maximum height of projectile motion can be obtained thus;

H = h +
`((v_(y ))^(2) )/(2g)`

H is the maximum height;

h is the height along the vertical h = 3m;

g is the acceleration due to gravity = 9.8 m/
`s^(2)`.

H = 3 +
`((21.9)^(2) )/(2 * 9.8)`

H = 27.4 m

The ball has a height of 27.4 m

5. The distance in reference to the green.

To know the reach, distance covered along the horizontal path and this can be obtained thus;

x =
`v_(x)` x t

x = 26.1 m/s x 4.6 s

x = 120.06 m

The distance is 120.06 m which is greater than 120 m

Since the distance is more than 120 m the golfer reached the green