Respuesta :
Answer:
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
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;
[tex]v_{y}[/tex] = vsinθ
[tex]v_{y}[/tex] is the vertical velocity
v is the initial velocity = 34 m/s
and θ is the projectile angle = [tex]40^{o}[/tex]
[tex]v_{y}[/tex] = (34 m/s) x (sin [tex]40^{o}[/tex])
[tex]v_{y}[/tex] = 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;
[tex]v_{x}[/tex] = vcosθ
[tex]v_{x}[/tex] is the horizontal velocity
v is the initial velocity = 34 m/s
and θ is the projectile angle = [tex]40^{o}[/tex]
[tex]v_{x}[/tex] = (34 m/s) x (cos [tex]40^{o}[/tex])
[tex]v_{x}[/tex] = 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 = [tex]v_{y}[/tex] t - [tex]\frac{1}{2}gt^{2}[/tex]
y is the vertical displacement (from the origin) = 0m- 3m = -3 m/s
[tex]v_{y}[/tex] is the initial vertical velocity = 21.9 m/s
g is the acceleration due to gravity = 9.8 m/[tex]s^{2}[/tex]
substituting the values in the expression.
3 = (21.9 x t) - ([tex]\frac{1}{2}[/tex] x 9.8 x [tex]t^{2}[/tex])
3 = 21.9 t - 4.9 [tex]t^{2}[/tex]
rearranging the equation we have;
4.9 [tex]t^{2}[/tex] - 21.9 t -3 = 0
using the quadratic formula to solve the equation we have;
t = 21.9 ± [tex]\frac{\sqrt{(-21.9) ^{2} -(4 * 4.9*(-3) )} }{2 * 4.9}[/tex]
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 + [tex]\frac{(v_{y })^{2} }{2g}[/tex]
H is the maximum height;
h is the height along the vertical h = 3m;
g is the acceleration due to gravity = 9.8 m/[tex]s^{2}[/tex].
H = 3 + [tex]\frac{(21.9)^{2} }{2 * 9.8}[/tex]
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 = [tex]v_{x}[/tex] 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
