A 0.25 kg ball is thrown vertically upward at a speed of 20 m / s. What is its kinetic energy at an altitude of 10m?

Ball motion equation:

h = V0 t – g t ^ 2/2.

where V0 is the initial speed of the ball, m / s;

g – acceleration due to gravity, g = 9.8 m / s2.

By the condition h = 10 m, we find the time t:

−4.9 t ^ 2 + 20 t – 10 = 0,

D = 20 ^ 2 – 4 × 4.9 × 10 = 204;

when lifting the ball up, we take the smaller root of the equation:

t1 = (20 – 204 ^ 0.5) / (2 × 4.9) = 0.583 s.

Speed at height h:

V1 = V0 – gt1 = 20 – 9.8 × 0.583 = 14.29 m / s.

Kinetic energy of the ball at height h:

Ec = M V1 ^ 2/2 = 0.25 × 14.29 ^ 2/2 = 25.5 J.

Answer: Ec = 25.5 J.