An Olympic diver is on a diving platform 10.1 m above the water. To start her dive, she runs off of the platform with a speed of 1.28 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water?

v = 14.13 m/s

Explanation:

We are given;

Height of platform above water; S_y = 10.1m

Horizontal component of velocity; V_x = 1.28 m/s

First of all, let's find time taken for the displacement.

t = √[(2S_y)/g]

t = √[(2 x 10.1)/9.8]

t = √[(20.2)/9.8]

t = 1.436 s

Let's use Newton's equation of motion for vertical component of velocity;

(v_y) = u + gt

(v_y) = 0 + (9. 8 x 1.436)

(v_y) = 14.07 m/s

So, final speed is the resultant of v_x and v_y.

So;

v = √((v_x)² + (v_y)²)

v = √((1.28)² + (14.07)²)

v = √199.6033

v = 14.13 m/s

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