A golf ball is chipped at an angle of 37 degrees and with a speed of 13.5 m/s. How far does it travel to the nearest tenth of a meter?
To find the distance traveled by the golf ball, we can analyze the horizontal and vertical components of its motion separately.
Given:
- Angle of 37 degrees
- Speed of 13.5 m/s
First, let's find the horizontal distance traveled by the golf ball.
1. Identify the horizontal component of the initial velocity: In this case, it is the initial velocity multiplied by the cosine of the angle.
Horizontal initial velocity = 13.5 m/s * cos(37°)
2. Calculate the time it takes for the golf ball to reach the maximum height. To do this, we need to find the time it takes for the vertical velocity to become zero.
Initial vertical velocity = 13.5 m/s * sin(37°)
Acceleration due to gravity = 9.8 m/s^2
Using the kinematic equation:
Final vertical velocity = Initial vertical velocity + (acceleration due to gravity * time)
0 = 13.5 m/s * sin(37°) + (-9.8 m/s^2 * time)
Solve for time:
time = (13.5 m/s * sin(37°)) / 9.8 m/s^2
3. Calculate the total time of flight. To do this, we double the time it took for the golf ball to reach the maximum height.
Total time of flight = 2 * time
4. Finally, calculate the horizontal distance traveled by the golf ball using the formula:
Horizontal distance = Horizontal initial velocity * Total time of flight
Plug in the values and solve for distance to the nearest tenth of a meter.