Question

A golf ball is hit with an initial velocity of 65 m/s at an angle of 36 degrees above the horizontal. How far does it travel?(horizontal distance)(g = 9.8 m/s²)

Answer 1

The horizontal distance traveled by the golf ball is approximately 185 meters.

Step-by-step explanation:

When a golf ball is hit, it travels in a trajectory that involves both horizontal and vertical components. To determine the horizontal distance traveled by the golf ball, we need to consider only the horizontal component of its motion, as the vertical component does not contribute to the horizontal distance.

Firstly, let's consider the initial velocity of the golf ball, which is given as 65 m/s at an angle of 36 degrees above the horizontal. We can decompose this initial velocity into its horizontal and vertical components using trigonometric functions:

`v_x` = v₀ sin(θ) = 65 m/s sin(36 degrees) = 41.8 m/s

`v_y` = v₀ cos(θ) = 65 m/s cos(36 degrees) = 47.2 m/s

Next, we can use the formula for distance traveled in one dimension (i.e., horizontal motion) to find the horizontal distance traveled by the golf ball:

x =
`v_x` t + x₀

Here, x represents the horizontal distance traveled,
`v_x` represents the horizontal component of initial velocity, t represents time taken to travel this distance, and x₀ represents the initial position of the golf ball (which is zero in this case). Since we are interested in finding the horizontal distance traveled, we can set t equal to the time taken for the ball to reach its maximum height (i.e., when y = 0). This time can be calculated using projectile motion formulas:

t = (2y₀)/g = (2 * (-47.2 m/s²) * (-19.3 m))/(9.8 m/s²) = 4.4 seconds

Now, we can substitute these values into our formula for horizontal distance:

x =
`v_x` t + x₀ = 41.8 m/s * 4.4 seconds + 0 meters = approximately 185 meters (rounded to two decimal places).