Task 1a

A rock is falling from the height 5,0 metres. At which speed does it hit the ground?

Solution: As we are not given the amount of time the rock spends in air, we have to use the law of conservation of energy. At the start, the rock has potential energy only, and at the moment of impact, it has kinetic energy only. Therefore:

[[$$E_{pot}=E_{kin} \quad$$]]
[[$$mgh = \frac{1}{2}mv^2 \quad | :m$$]]
[[$$gh = \frac{1}{2}v^2 \quad | \cdot 2$$]]
[[$$2gh = v^2 \quad$$]]
[[$$v^2 = 2gh \quad | \sqrt{ }$$]]
[[$$v=\sqrt{2gh}=\sqrt{2\cdot 10 \frac{m}{s^2}\cdot 5,0 m}=10 \frac{m}{s}$$]]

Task 1b

A rock is falling from a certain height for 5,0 seconds. At which speed does it hit the ground?

Solution: As we are given the amount of time the rock spends in air, we may use the formula for acceleration. Therefore:

[[$$a=\frac{\Delta v}{\Delta t}\quad |\cdot \Delta t$$]]
[[$$\Delta v = a \cdot \Delta t=10 \frac{m}{s^2}\cdot 5,0 s = 50 \frac{m}{s}$$]]

Task 2a

You throw a ball upwards at [[$5,0 \frac{m}{s}$]]. How high does it go?

Solution: As we are not given the amount of time the ball spends in air, we have to use the law of conservation of energy. At the start, the ball has potential energy only, and at the moment of impact, it has kinetic energy only. Therefore:

[[$$E_{pot}=E_{kin} \quad$$]]
[[$$mgh = \frac{1}{2}mv^2 \quad | :m$$]]
[[$$gh = \frac{1}{2}v^2 \quad | :h$$]]
[[$$h = \frac{v^2}{2g}=\frac{(5 \frac{m}{s})^2}{2\cdot 10\frac{m}{s^2}}=1,25 m\approx 1,3 m$$]]

Task 2b

You throw a ball upwards at [[$5,0 \frac{m}{s}$]]. How long does it fly?

Solution: As we are given the amount of time the ball spends in air, we may use the formula for acceleration. Because the velocity of the ball at the start is [[$5,0 \frac{m}{s}$]] upwards and then at the end [[$5,0 \frac{m}{s}$]] downwards, the change in velocity [[$\Delta v$]] is [[$10\frac{m}{s}$]] Therefore:

[[$$a=\frac{\Delta v}{\Delta t}\quad |\cdot \Delta t : a$$]]
[[$$\Delta t = \frac{\Delta v}{a}= \frac{10\frac{m}{s}}{10\frac{m}{s^2}}=1,0 s$$]]