A golfer hits a golf ball with a pitching wedge from a T that is 15 m lower than the green if the ball leaves tea at 35 m/s at an angle of 40° to the horizontal and it lands in the hole on the green what is the horizontal distance from the golfer to the whole the answer you should get is 101.3 m
To solve this problem, we need to use the kinematic equations of motion in two dimensions.
Given:
Initial velocity (v0) = 35 m/s
Launch angle (θ) = 40°
Vertical displacement (Δy) = -15 m (negative because the ball is moving downward)
Horizontal distance (Δx) = ?
Acceleration due to gravity (g) = -9.81 m/s^2
We can find the time of flight by using the vertical displacement equation:
Δy = v0sin(θ)t - (1/2)gt^2
-15 = 35sin(40°)t - (1/2)(9.81)t^2
-15 = 22.6t - 4.905t^2
Rearranging the equation, we get:
4.905t^2 - 22.6t - 15 = 0
Using the quadratic formula, we find that t ≈ 3.062 seconds.
Now, we can find the horizontal distance using the horizontal displacement equation:
Δx = v0cos(θ)t
Δx = 35cos(40°) * 3.062
Δx ≈ 101.3 m
Therefore, the horizontal distance from the golfer to the hole is approximately 101.3 meters.