# The athlete jumped from an eight-meter platform and plunged into the water to a depth of 2.5 meters.

**The athlete jumped from an eight-meter platform and plunged into the water to a depth of 2.5 meters. With which accelerated and for how long did it move in the water to a stop? Neglect air resistance.**

h = 8 m.

g = 10 m / s ^ 2.

l = 2.5 m.

a -?

t -?

Since the athlete jumps without an initial speed and under the action of only gravity, the athlete’s speed V at the moment of entering the water is expressed from the formula:

h = V ^ 2/2 * g.

V ^ 2 = 2 * g * h.

The athlete’s starting speed in the water will be V, and the athlete’s final speed will be 0.

The depth of its immersion l will be determined by the formula: l = V ^ 2/2 * a, where a is its acceleration in the water.

a = V ^ 2/2 * l = 2 * g * h / 2 * l = g * h / l.

a = 10 m / s ^ 2 * 8 m / 2.5 m = 32 m / s ^ 2.

t = V / a = √ (2 * g * h) / a.

t = √ (2 * 10 m / s ^ 2 * 8 m) / 2.5 m = 5.05 s.

Answer: a = 32 m / s ^ 2, t = 5.05 s.