Final answer:
To find the initial velocity magnitude and launch angle of the golf ball, we calculate the horizontal velocity component and use it along with the time of flight to determine the vertical velocity component. Solving for these gives us the magnitude of the initial velocity. We then use trigonometry to find the launch angle.
Step-by-step explanation:
To solve this problem, we need to use equations for projectile motion. Given that the golf ball lands 92.2m away after 4.3 seconds, we can use the horizontal motion equation to find the initial horizontal velocity (vx).
We begin by calculating the horizontal velocity using the formula vx = d / t, where d is the total horizontal distance and t is the time. Substituting the given values:
vx = 92.2m / 4.3s = 21.44 m/s
Now, interestingly, the initial horizontal velocity (vx) is also v0cos(θ), where v0 is the initial velocity and θ is the launch angle. To solve for the initial velocity, we need the vertical velocity component. The vertical velocity at the point of landing vy = 0 because at the peak of the trajectory the vertical velocity is zero (disregarding the second half of the trajectory).
v0y = v0sin(θ). Given that the acceleration due to gravity g = 9.8 m/s², the time to reach the peak is t = v0y / g. We know that the time to reach the peak is half of the total time, thus the equation becomes 4.3s / 2 = v0sin(θ) / 9.8 m/s².
Combining the horizontal and vertical components, we have:
- vx = v0cos(θ)
- 2.15s = v0sin(θ) / 9.8 m/s²
Squaring both equations and adding them eliminates θ, giving us:
v0² = vx² + (v0y)²
Substituting vx and solving for v0y enables us to calculate the initial velocity and then use it to find the launch angle θ. With both components known, we can use trigonometry to obtain the angle.
v0 = sqrt(vx² + (v0y)²)
Finding θ:
θ = atan(v0y/vx)
After computing the calculations, you find the initial velocity magnitude and angle.