A golfer, standing on a fairway, hits a shot to a green that is elevated 5.50 m above the point where she is standing. The ball leaves her club at an angle of 37.0° above the ground, with a speed of 31.1 m/s.
Find the time that the ball is in the air before it hits the green.
How far did the ball travel horizontally?
Determine the ball's speed right before it hits the green.
Vo = 31.1m/s[37o]
Xo = 31.1*Cos37 = 24.84 m/s.
Yo = 31.1*sin37 = 18.72 m/s.
Y = Yo + g*Tr = 0
18.72 - 9.8*Tr = 0
9.8Tr = 18.72
Tr = 1.91 s. = Rise time.
Y^2 = Yo^2 + 2g*h = 0
h=-Yo^2/2g = -(18.72)^2/-19.6 = 17.88 m
= Max. ht.
0.5g*t^2 = 17.88 - 5.50
4.9t^2 = 12.38
t^2 = 2.53
t = 1.59 s. = Fall time(Tf).
a. Tr+Tf = 1.91 + 1.59 = 3.50 s. = Time in air.
b. D = Xo*(Tr+Tf) = 24.84 * 3.50 = 86.9 m.
c. Y = Yo + g*Tf = 0 + 9.8*1.59 = 15.58 m/s. = Ver. component of final velocity.
V = sqrt(Xo^2+Y^2)
To find the time that the ball is in the air before it hits the green, we can use the kinematic equation for vertical motion:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
y = final vertical position (5.50 m)
y0 = initial vertical position (0 m)
v0y = initial vertical velocity (v0 * sin(θ))
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Solving for time, we get:
5.50 = 0 + (31.1 * sin(37°)) * t - (1/2) * (-9.8) * t^2
Simplifying the equation, we have:
5.50 = 31.1 * sin(37°) * t + 4.9 * t^2
Rearranging the equation gives us a quadratic equation in the form of at^2 + bt + c = 0:
4.9 * t^2 + 31.1 * sin(37°) * t - 5.50 = 0
We can solve this quadratic equation to find the value of t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values, we have:
t = (-(31.1 * sin(37°)) ± √((31.1 * sin(37°))^2 - 4 * 4.9 * (-5.50))) / (2 * 4.9)
Calculating this expression, we find that the positive value of t is approximately 1.43 seconds.
Therefore, the ball is in the air for approximately 1.43 seconds before it hits the green.
To find how far the ball traveled horizontally, we can use the kinematic equation for horizontal motion:
x = x0 + v0x * t
Where:
x = distance traveled horizontally
x0 = initial horizontal position (0 m)
v0x = initial horizontal velocity (v0 * cos(θ))
t = time
Since the ball starts with an initial horizontal position of 0 m and there is no acceleration horizontally, the equation simplifies to:
x = v0x * t
Plugging in the values, we have:
x = (31.1 * cos(37°)) * 1.43
Calculating this expression, we find that the ball traveled approximately 25.71 meters horizontally.
Finally, to determine the ball's speed right before it hits the green, we can use the equation for velocity:
v = √(v0x^2 + v0y^2)
Where:
v = final velocity
v0x = initial horizontal velocity (v0 * cos(θ))
v0y = initial vertical velocity (v0 * sin(θ))
Plugging in the values, we have:
v = √((31.1 * cos(37°))^2 + (31.1 * sin(37°))^2)
Calculating this expression, we find that the ball's speed right before it hits the green is approximately 31.1 m/s.