Divers in Acapalco dive from a cliff that is 60 m high. If the rocks below the cliff extend outward for 26 m, what is the minimum horizontal velocity a diver must have to clear the rocks?

To find the minimum horizontal velocity a diver must have to clear the rocks, we can use the principles of projectile motion.

Let's assume the initial velocity of the diver is composed of two components: horizontal velocity (Vx) and vertical velocity (Vy).

The vertical motion of the diver can be analyzed separately, using the equation for motion with constant acceleration:

h = (Vy * t) + (0.5 * g * t^2),

where h is the height of the cliff (60 m), Vy is the vertical velocity, t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

We can rearrange the equation to solve for t:

t = sqrt((2 * h) / g).

Since the horizontal distance traveled (x) is given as 26 m, we can calculate the time of flight:

t = x / Vx.

Setting these two equations equal, we can solve for Vx:

sqrt((2 * h) / g) = x / Vx.

Simplifying, we get:

Vx = (x / sqrt((2 * h) / g)).

Substituting the given values, we have:

Vx = (26 / sqrt((2 * 60) / 9.8)).

Evaluating this equation will give us the minimum horizontal velocity a diver must have to clear the rocks.

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