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A golf ball is hit so that it travels a horizontal distance of 440 feet and reaches a maximum height of 190 feet.

User David Dury
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We can use the following kinematic equations to solve the problem:

For vertical motion:
v_f^2 = v_i^2 + 2ad
where v_f is the final velocity (0 m/s at maximum height), v_i is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and d is the displacement (190 feet or 58.0 meters).

From this equation, we can solve for v_i:
v_i = sqrt(v_f^2 - 2ad)

For horizontal motion:
d = v_avg * t
where d is the displacement (440 feet or 134.1 meters), v_avg is the average velocity (which is equal to the horizontal component of the initial velocity), and t is the time of flight.

From this equation, we can solve for the initial velocity:
v_i = d / t

We can now equate the expressions for v_i and solve for t:

sqrt(v_f^2 - 2ad) = d / t

t = d / sqrt(v_f^2 - 2ad)

Plugging in the values, we get:

t = 134.1 m / sqrt((0 m/s)^2 - 2(-9.8 m/s^2)(58.0 m))
t = 4.17 s

Now we can use the equation for horizontal motion to find the initial velocity:

v_i = d / t
v_i = 134.1 m / 4.17 s
v_i = 32.2 m/s

Therefore, the initial velocity of the golf ball was 32.2 m/s.
User Dhruvil Patel
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