Question

A ball is thrown up into the air. When it falls back down and reaches the

ground, its final downward velocity is 8.0 m/s. What was the maximum height
of the ball? (Hint: KEinitial = 0, PEfinal = 0.)
O A. 3.3 m
O B. 6.5 m
O c. 0.41
O D. 32 m

Answer 1

A. 3.3 m

Step-by-step explanation:

Here, we have to use conservation of energy principle.

When the ball is at maximum height, the instantaneous velocity at that point is 0 m/s. So, the kinetic energy of the ball is also 0 at the maximum height. Thus, at maximum height, the energy possessed by the ball is gravitational potential energy only.

Now, when the ball reaches the ground, all the gravitational potential energy changes into kinetic energy because of the conservation of energy.

Therefore, the energy transformation can be given as:

Decrease in potential energy = Increase in Kinetic energy

Decrease in potential energy is given as:

`\Delta U=mg(h-0)=mgh`

Increase in kinetic energy is given as:

`\Delta K=(1)/(2)mv^2-0=(1)/(2)mv^2`

Therefore,

`\Delta U=\Delta K\\mgh=(1)/(2)mv^2\\h=(v^2)/(2g)`

Now, plug in 8 for
`v`, 9.8 for
`g` and solve for height
`h`. This gives,

`h=(8^2)/(2* 9.8)\\h=(64)/(19.6)=3.265\approx 3.3\ m`

Therefore, the maximum height reached by the ball is 3.3 m.