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. If the ball leaves her club with a velocity of 46.0 m/s at an angle 0f 35.0º above the ground, find the time that the ball is in the air before it hits the green.
Vi = 46 sin 35 = 26.4 m/s up
5.5 = 0 + 26.4 t - 4.9 t^2
4.9 t^2 -26.4 t + 5.5 = 0
solve quadratic, use the longer time
To find the time that the ball is in the air before it hits the green, we can use the equations of motion.
Step 1: Split the initial velocity into horizontal and vertical components:
The horizontal component of the initial velocity is given by Vx = V * cos(θ) = 46.0 m/s * cos(35.0º)
The vertical component of the initial velocity is given by Vy = V * sin(θ) = 46.0 m/s * sin(35.0º)
Step 2: Find the time of flight:
The time of flight, T, is the total time the ball is in the air before it hits the ground.
Using the vertical component, we can find the time it takes for the ball to reach its highest point:
Vy = V * sin(θ) = g * t_up
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t_up is the time taken to reach the highest point.
t_up = Vy / g
At the highest point, the vertical velocity becomes zero.
Therefore, the total time in the air is twice the time taken to reach the highest point.
T = 2 * t_up
Step 3: Substitute the given values and calculate:
Vy = 46.0 m/s * sin(35.0º)
t_up = (46.0 m/s * sin(35.0º)) / 9.8 m/s^2
T = 2 * ((46.0 m/s * sin(35.0º)) / 9.8 m/s^2)
Calculate each step:
Vy = 46.0 m/s * sin(35.0º)
Vy = 46.0 m/s * 0.5745
Vy ≈ 26.473 m/s
t_up = (46.0 m/s * sin(35.0º)) / 9.8 m/s^2
t_up = (26.473 m/s) / 9.8 m/s^2
t_up ≈ 2.703 seconds
T = 2 * ((46.0 m/s * sin(35.0º)) / 9.8 m/s^2)
T = 2 * (26.473 m/s / 9.8 m/s^2)
T ≈ 5.406 seconds
Therefore, the ball is in the air for approximately 5.406 seconds before it hits the green.
To find the time that the ball is in the air before it hits the green, we can use the vertical motion equation:
h = v₀t + (1/2)gt²
Where:
h = vertical displacement of the ball (in this case, 5.50 m)
v₀ = initial vertical velocity of the ball (in this case, the vertical component of the initial velocity)
t = time the ball is in the air
g = acceleration due to gravity (approximately 9.8 m/s²)
First, we need to find the initial vertical velocity of the ball, v₀. To do that, we can use the trigonometric relationship between the given initial velocity (46.0 m/s) and the angle (35.0º) above the ground:
v₀ = v * sin(θ)
v₀ = 46.0 m/s * sin(35.0º)
v₀ ≈ 26.20 m/s
Now, we can substitute the known values into the vertical motion equation:
5.50 m = 26.20 m/s * t + (1/2) * 9.8 m/s² * t²
Simplifying the equation:
5.50 m = 26.2 m/s * t + 4.9 m/s² * t²
Rearranging the equation into the quadratic form:
4.9 m/s² * t² + 26.20 m/s * t - 5.50 m = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In this case:
a = 4.9 m/s²
b = 26.20 m/s
c = -5.50 m
Substituting these values into the quadratic formula:
t = (-26.20 m/s ± √((26.20 m/s)² - 4 * 4.9 m/s² * (-5.50 m))) / (2 * 4.9 m/s²)
Calculating the square root and simplifying further:
t = (-26.20 m/s ± √(686.44 m² + 107.80 m²)) / 9.8 m/s²
t = (-26.20 m/s ± √(794.24 m²)) / 9.8 m/s²
t = (-26.20 m/s ± 28.17 m) / 9.8 m/s²
Now, we can calculate the two possible values for t:
t₁ = (-26.20 m/s + 28.17 m) / 9.8 m/s²
t₁ ≈ 0.21 s
t₂ = (-26.20 m/s - 28.17 m) / 9.8 m/s²
t₂ ≈ -5.38 s (discarded since time cannot be negative)
Therefore, the time that the ball is in the air before it hits the green is approximately 0.21 seconds.