A 0.5 ext kg object is thrown upwards with a velocity of 18 ext m/s. Determine the height at which potential energy equals kinetic energy.

The total mechanical energy is conserved. At the point of projection (height $h=0$), all energy is kinetic.

$$E_{total} = KE_0 = \frac{1}{2}mv_0^2$$ $$E_{total} = \frac{1}{2}(0.5 \text{ kg})(18 \text{ m/s})^2$$ $$E_{total} = \frac{1}{2}(0.5 \text{ kg})(324 \text{ m}^2/\text{s}^2)$$ $$E_{total} = 81 \text{ J}$$

d) The mechanical energy is the total energy, which is conserved.

$$\boxed{\text{Mechanical energy} = 81 \text{ J}}$$

At this height $h$, $PE = KE$. Since $E_{total} = KE + PE$, we have $E_{total} = PE + PE = 2PE$ or $E_{total} = KE + KE = 2KE$.
So, $PE = \frac{1}{2}E_{total}$ and $KE = \frac{1}{2}E_{total}$.

$$PE = mgh$$ $$\frac{1}{2}E_{total} = mgh$$ $$\frac{1}{2}(81 \text{ J}) = (0.5 \text{ kg})(9.8 \text{ m/s}^2)h$$ $$40.5 \text{ J} = (4.9 \text{ N})h$$ $$h = \frac{40.5 \text{ J}}{4.9 \text{ N}}$$ $$h \approx 8.27 \text{ m}$$

a) The height at which potential energy equals kinetic energy is:

$$\boxed{h = 8.27 \text{ m}}$$

b) At this height, the kinetic energy is half of the total mechanical energy.

$$KE = \frac{1}{2}E_{total} = \frac{1}{2}(81 \text{ J})$$ $$\boxed{KE = 40.5 \text{ J}}$$

c) At this height, the potential energy is also half of the total mechanical energy.

$$PE = \frac{1}{2}E_{total} = \frac{1}{2}(81 \text{ J})$$ $$\boxed{PE = 40.5 \text{ J}}$$

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: A ball is projected upward. At a certain height, its speed is $12 \text{ m/s}$ and its kinetic energy equals its potential energy. Calculate the height at that point.

Given:
Speed $v = 12 \text{ m/s}$ at a certain height $h$.
At this height, $KE = PE$.
Acceleration due to gravity $g = 9.8 \text{ m/s}^2$.

$$KE = \frac{1}{2}mv^2$$ $$PE = mgh$$

$$\frac{1}{2}mv^2 = mgh$$

Cancel out the mass $m$ from both sides:

$$\frac{1}{2}v^2 = gh$$

$$h = \frac{v^2}{2g}$$

Substitute the given values:

$$h = \frac{(12 \text{ m/s})^2}{2(9.8 \text{ m/s}^2)}$$ $$h = \frac{144 \text{ m}^2/\text{s}^2}{19.6 \text{ m/s}^2}$$ $$h \approx 7.35 \text{ m}$$

The height at that point is:

$$\boxed{h = 7.35 \text{ m}}$$