A man weighing 70 kg jumped from a high bank into the sea. What is the speed of a person’s movement 0.5 s after a jump? The influence of air resistance is not taken into account, the force of gravity acting on a person is 700 N.
m = 70 kg.
t = 0.5 s.
V0 = 0 m / s.
g = 10 m / s ^ 2.
F = 700 N.
Method 1: kinematic.
The body moves with the acceleration of gravity g. The speed of movement V is determined by the formula: V = V0 + g * t. Since V0 = 0 m / s, then V = g * t.
V = 10 m / s ^ 2 * 0.5 s = 5 m / s.
Method 2: dynamic.
Let’s write Newton’s 2 law for a body: m * a = F.
a = F / m.
Let’s write down the formula for determining the acceleration: a = (V – V0) / t.
Since V0 = 0 m / s, then a = V / t.
V / t = F / m.
V = F * t / m.
V = 700 N * 0.5 s / 70 kg = 5 m / s.
Answer: the speed of the body will be V = 5 m / s.
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