a golfer hits a golf ball an initial velocity of 54m/s [42 degrees above the horizontal]. the ball lands on an elevated green that is 14m higher that the level from which the golfer hit the ball.
a) how far away horizontally from the golfer does the ball land?
b) what is the velocity of the ball when it lands?
please show all the steps
Vo = 54m/s[42o]
Xo = 54*Cos42 = 40.1 m/s.
Yo = 54*sin42 = 36.1 m/s.
a. Y = Yo + g*Tr = 0
Tr = -Yo/g = -36.1/-9.8 = 3.69 s. = Rise
time.
Y^2 = Yo^2 + 2g*h = 0
h = -(Yo^2)/2g = (36.1^2)/-19.6=66.5 m.
0.5g*t^2 = 66.5-14 = 52.5
4.9t^2 = 52.5
t^2 = 10.71
t = 3.27 s. = Fall time(Tf).
Dx = Xo * (Tr+Tf) = 40.1 * (3.69+3.27) =
279.1 m. = Hor. distance.
b. Y^2 = Yo^2 + 2g*h
Y = Ver. component of final velocity.
Yo = 0
g = +9.8 m/s^2
h = 66.5-14 = 52.5 m.
Solve for Y.
V = sqrt(Xo^2+Y^2) m/s
To find the distance the golf ball lands from the golfer, we need to break down the initial velocity into horizontal and vertical components.
Let's start with the horizontal component. The initial velocity can be resolved into horizontal and vertical components using trigonometry. The horizontal component will be the initial velocity multiplied by the cosine of the launch angle.
a) Horizontal distance:
Horizontal component = Initial velocity * cos(angle)
In this case,
Initial velocity = 54 m/s
Angle = 42 degrees
Horizontal component = 54 m/s * cos(42 degrees)
Using a calculator, the horizontal component is approximately 54 m/s * 0.7431 = 40.08 m/s.
Now, we can move on to the vertical component of the initial velocity. The vertical component will be the initial velocity multiplied by the sine of the launch angle.
Vertical component = Initial velocity * sin(angle)
Vertical component = 54 m/s * sin(42 degrees)
Using a calculator, the vertical component is approximately 54 m/s * 0.6691 = 36.10 m/s.
Next, we need to find the time it takes for the ball to reach the elevated green. We can use the equation for vertical motion:
Vertical displacement = Vertical component * time - (1/2) * acceleration * time^2
Since the ball lands on the elevated green, its vertical displacement is -14 m (negative because it is a downward displacement), acceleration is -9.8 m/s^2 (due to gravity), and we want to find the time. Rearrange the equation:
0 = (1/2) * (-9.8 m/s^2) * time^2 + (36.10 m/s) * time - 14 m
Solve this quadratic equation for time. In this case, we will use the quadratic formula:
time = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
time = [-(36.10 m/s) ± √((36.10 m/s)^2 - 4 * (1/2) * (-9.8 m/s^2) * (-14 m)] / [2 * (1/2) * (-9.8 m/s^2)]
Using a calculator, you will find two solutions for time: t1 ≈ 0.79 s and t2 ≈ 5.78 s. Since the ball is still in the air after 0.79 s, we will use the longer time, t2 = 5.78 s.
Now, we can find the horizontal distance the ball travels:
Horizontal distance = Horizontal component * time
Horizontal distance = (40.08 m/s) * (5.78 s)
Using a calculator, the horizontal distance is approximately 231.57 m.
Therefore, the ball lands approximately 231.57 meters horizontally from the golfer.
b) To find the velocity of the ball when it lands, we need to find the final velocity in the vertical direction (Vfy) and the final velocity in the horizontal direction (Vfx).
Vfy = Vertical component - acceleration * time
Vfy = (36.10 m/s) - (-9.8 m/s^2) * (5.78 s)
Vfy ≈ 86.05 m/s
Vfx = Horizontal component
Vfx = 40.08 m/s
The velocity of the ball when it lands is the vector sum of the horizontal and vertical components:
Velocity = sqrt(Vfx^2 + Vfy^2)
Velocity = sqrt((40.08 m/s)^2 + (86.05 m/s)^2)
Using a calculator, the velocity of the ball when it lands is approximately 95.71 m/s.