A golfer drives a golf ball with a velocity of 100 ft/sec and an angle of 45 degrees. A. find the time of travel for the ball. B. Find the balls range. C. Find the maximum height of the ball
67yards
To answer these questions, we can use the principles of projectile motion. Projectile motion refers to the motion of an object that is launched into the air and is subject to only the force of gravity and air resistance (neglected in this case). In this scenario, the golf ball can be treated as a projectile.
Let's break down each question and solve them step by step:
A. To find the time of travel for the ball, we can use the horizontal component of the initial velocity. The given angle of 45 degrees splits the initial velocity into equal horizontal and vertical components. Each component can be calculated using trigonometry.
The horizontal component (Vx) can be found using the formula:
Vx = V * cos(θ)
where V is the initial velocity (100 ft/sec) and θ is the angle of launch (45 degrees).
Substituting the values into the formula:
Vx = 100 ft/sec * cos(45 degrees)
= 100 ft/sec * (√2/2)
= 100 ft/sec * 0.7071
≈ 70.71 ft/sec
Now, to find the time of travel, we can divide the horizontal distance traveled by the horizontal component of velocity. Since there is no horizontal acceleration, this will give us the time.
Distance traveled (horizontal) = Range = R
Thus:
R = Vx * t
t = R / Vx
B. To find the ball's range, we can use the formula:
Range (R) = Vx * t
We already found Vx in the previous step (70.71 ft/sec), and we'll use t, the time of travel, which we can calculate using the equation derived above.
C. To find the maximum height of the ball, we need to determine the vertical component of the initial velocity (Vy). We can use the formula:
Vy = V * sin(θ)
Substituting the values into the formula:
Vy = 100 ft/sec * sin(45 degrees)
= 100 ft/sec * (√2/2)
= 100 ft/sec * 0.7071
≈ 70.71 ft/sec
The maximum height (H) can be found using the following formula:
H = (Vy^2) / (2g)
where g is the acceleration due to gravity (32.2 ft/sec^2).
Substituting the values into the formula:
H = (70.71 ft/sec)^2 / (2 * 32.2 ft/sec^2)
= 5000 ft^2/sec^2 / 64.4 ft/sec^2
≈ 77.6 ft
So, the maximum height of the ball is approximately 77.6 ft under these given conditions.