A ball of mass 0.1 kg is thrown upwards with an initial velocity of 80 m/s. Calculate its kinetic energy as it leaves the ground, its potential energy at maximum height, and its potential energy halfw

Here's the solution for question 20:

Given:

• Mass of the ball, $m = 0.1\,kg$
• Initial velocity, $u = 80\,m/s$
• Acceleration due to gravity, $g = 9.8\,m/s^2$

Part (iii): Kinetic energy as it leaves the ground.
The kinetic energy as it leaves the ground is its initial kinetic energy.
The formula for kinetic energy is $KE = \frac{1}{2}mv^2$.

$$KE_{initial} = \frac{1}{2}mu^2$$ $$KE_{initial} = \frac{1}{2}(0.1\,kg)(80\,m/s)^2$$ $$KE_{initial} = \frac{1}{2}(0.1)(6400)$$ $$KE_{initial} = 0.05 \times 6400$$ $$KE_{initial} = 320\,J$$

The kinetic energy as it leaves the ground is $\boxed{\text{320 J}}$.

Step 1: Calculate the maximum height ($H_{max}$).
At the maximum height, the final velocity ($v$) of the ball is $0\,m/s$. Using the kinematic equation $v^2 = u^2 + 2as$, where $a = -g$ (since gravity acts downwards) and $s = H_{max}$:

$$0^2 = u^2 + 2(-g)H_{max}$$ $$0 = (80\,m/s)^2 - 2(9.8\,m/s^2)H_{max}$$ $$0 = 6400 - 19.6 H_{max}$$ $$19.6 H_{max} = 6400$$ $$H_{max} = \frac{6400}{19.6}$$ $$H_{max} \approx 326.53\,m$$

Part (ii): Potential energy at its maximum height.
The formula for potential energy is $PE = mgh$.

$$PE_{max} = mgH_{max}$$ $$PE_{max} = (0.1\,kg)(9.8\,m/s^2)(326.53\,m)$$ $$PE_{max} = 320\,J$$

The potential energy at its maximum height is $\boxed{\text{320 J}}$.

Part (i): Potential energy half way up.
Half way up, the height is $H_{half} = \frac{H_{max}}{2}$.

$$H_{half} = \frac{326.53\,m}{2}$$ $$H_{half} \approx 163.265\,m$$

Now, calculate the potential energy at this height:

$$PE_{half} = mgH_{half}$$ $$PE_{half} = (0.1\,kg)(9.8\,m/s^2)(163.265\,m)$$ $$PE_{half} = 160\,J$$

The potential energy half way up is $\boxed{\text{160 J}}$.