A golfer drives a ball with a velocity of 42 m/s at an angle of 45 degrees above the horizontal. What is the maximum height (in meters) of the ball above the level fairway?
figure vertical velocity intial: 42sin45
then, knowing the vertical velocity at the peak is zero...
Vf^2=Vi^2+2gd
figure d
so does that mean vf is zero, vi is 29.69848481, and g is 9.8
g is -9.8m/s^2, so the answer d will be postive, or it should be. Your carrying all those digits on V is nonsense. You are restricted to two significant digits, given the info in the problem.
To find the maximum height, we can use the equations of motion for projectile motion.
First, let's break down the initial velocity of the ball into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes due to the gravitational force.
The initial velocity in the x-direction (horizontal) is given by:
Vx = V * cos(theta)
where V is the magnitude of the initial velocity (42 m/s) and theta is the launch angle (45 degrees).
The initial velocity in the y-direction (vertical) is given by:
Vy = V * sin(theta)
At the highest point of the ball's trajectory, the vertical velocity component will be zero, and only the horizontal velocity component will contribute to the motion.
To find the time it takes for the ball to reach its maximum height, we can use the equation:
Vy = V0y - g * t
where V0y is the initial vertical velocity (Vy), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Since the ball reaches its maximum height when Vy = 0, we can solve for t:
0 = Vy - g * t
t = Vy / g
Now we can find the time it takes for the ball to reach its maximum height:
t = (V * sin(theta)) / g
The maximum height of the ball can be found using the equation:
h = V0y * t - (1/2) * g * t^2
where h is the maximum height.
Substituting the values we have:
h = (V * sin(theta)) * [(V * sin(theta)) / g] - (1/2) * g * [(V * sin(theta)) / g]^2
h = (V^2 * sin^2(theta)) / (2 * g)
Plugging in the given values:
h = (42^2 * sin^2(45)) / (2 * 9.8)
Calculating further, we get:
h = (1764 * 0.5) / 19.6
h = 88.2 / 19.6
h ≈ 4.49 meters
Therefore, the maximum height of the ball above the level fairway is approximately 4.49 meters.