A golfer hits a golf ball, giving it an initial speed of 31.7 m/s at an angle of 55° above the horizontal. How far in meters will the golf ball travel horizontally before hitting the ground?

Vo = 31.7m/s[55o]

Xo = 31.7*cos55 = 18.18 m/s.
Yo = 31.7*sin55 = 25.97 m/x.

Dx = Vo^2*sin(2A)/g=31.7^2*sin(110)/9.8
= 96.4 m.

what is the value of A in this equation

To find the horizontal distance traveled by the golf ball before hitting the ground, we can use the kinematic equations of motion.

First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component of the initial velocity (v₀x) can be found using the equation:

v₀x = v₀ * cos(θ)

where v₀ is the initial speed (31.7 m/s) and θ is the launch angle (55°).

v₀x = 31.7 m/s * cos(55°)
v₀x ≈ 31.7 m/s * 0.5736
v₀x ≈ 18.2 m/s

The vertical component of the initial velocity (v₀y) can be found using the equation:

v₀y = v₀ * sin(θ)

v₀y = 31.7 m/s * sin(55°)
v₀y ≈ 31.7 m/s * 0.8192
v₀y ≈ 25.9 m/s

Now, we can find the time it takes for the ball to hit the ground using the equation:

t = 2 * v₀y / g

where g is the acceleration due to gravity (approximately 9.8 m/s²).

t = 2 * 25.9 m/s / 9.8 m/s²
t ≈ 5.28 s

Finally, we can calculate the horizontal distance (d) traveled by the golf ball using the equation:

d = v₀x * t

d = 18.2 m/s * 5.28 s
d ≈ 96.1 m

Therefore, the golf ball will travel approximately 96.1 meters horizontally before hitting the ground.

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