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.