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If a 0.2-kg ball is dropped from a height of 10 m, what is its kinetic energy when it hits the ground?

User Anji R
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1 Answer

3 votes

Answer:19.6J

Step-by-step explanation:

There are two ways of solving this problem.

1st, in terms of kinematics we can solve for final velocity and plug into the kinetic energy equation 1/2 mv^2.

Solve for final velocity

The ball will have a displacement of 10 m downward or -10 m

The initial velocity is 0

The acceleration is g = -9.8 m/s^2

We plug into the equation V_f^2=V_i^2+2(a)(displacement)

v_f^2= 0+2(-9.8)(-10)

vf=14 m/s

Now plug into Ke=(1/2)m(V^2)

Ke = (1/2)0.2(14^2)= 19.6 J

The second way we can solve is using conservation of energy

Gravitational potential energy (mgh) at max height equals Kinetic energy (1/2)mv^2at the ground.

mgh=1/2mv^2 =Ke

.2*9.8*10=Ke= 19.6 J

User ThangLe
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