What is the kinetic energy of a body with a mass of m = 200 g, thrown horizontally from a height of h = 100 m

| December 16, 2020 | education

What is the kinetic energy of a body with a mass of m = 200 g, thrown horizontally from a height of h = 100 m with a speed of modulus v = 30 m / s after a time t = 2 s after throwing.

The kinetic energy of uniformly accelerated motion is found by the formula
Ek = mV ^ 2/2-mVo ^ 2/2
We have everything except the final and initial speed, in order to find it we need to derive the initial speed from the formula
V = Vo + at
And when we find it, we simply substitute it into this formula and derive the final speed
S = V ^ 2-Vo ^ 2 / 2a
Now we substitute the values
30 = Vo + 10 * 2
Vo = 30-20 = 10 (m / s)
Now we substitute the values ​​of the initial speed into the formula for the path and find the final speed
100 = V ^ 2-100 / 20
There comes out a quadratic equation and
V = 45.8
Substitute and find kinetic energy
Ek = 0.2 * 45.8 ^ 2 / 2-0.2 * 10 ^ 2/2
Ek = 199.764 (J)
Answer: Ek = 199.764 (J)



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