### Lunar Leaper

Gravity on the Moon is about 1/6th that on the Earth. A pole-vaulter 2 metres tall can clear a 5 metres pole on the Earth. How high a pole could he clear on the Moon?

### High Jumping

How high can a high jumper jump? How can a high jumper jump higher without jumping higher? Read on...

### Escape from Planet Earth

How fast would you have to throw a ball upwards so that it would never land?

# Slingshot

##### Stage: 5 Challenge Level:

We use conservation of energy to find the speed of the stone. All the energy stored in the elastic bands is transferred to the stone as kinetic energy.

Energy stored in the elastic bands is $$E =2 \frac{k(\sqrt{x^2 + (\sqrt{L^2 - H^2})^2} - \sqrt{L^2 - H^2})^2}{2}$$ because we have two of them and the amount by which one is extended is $$\sqrt{x^2 + (\sqrt{L^2 - H^2})^2} - \sqrt{L^2 - H^2}$$

Suppose that the initial speed of the stone is $v$ then it has a kinetic energy $E = \frac{mv^2}{2}$. The stone is fired at an angle of 45°. The initial speed of a projectile and the distance it travels is related by $d = \frac{v^2}{g}$. It is given that $d = 50$m. Thus, the kinetic energy is $E = \frac{mgd}{2}$.

By the conservation of energy $${k(\sqrt{x^2 + L^2 - H^2} - \sqrt{L^2 - H^2})^2} = \frac{mgd}{2}$$ Now, the numbers $L = 0.12$m , $H = 0.10$ m , $m = 0.05$kg , $k = 200$N/m , $d = 50$ m, $g = 9.81$ m/s^2 could be plugged in to find $x$ or we can simplify to get $$x = \sqrt{\frac{mgd}{2k} + \sqrt{(L^2 - H^2) \frac{2mgd}{k}}}$$

Thus, we have that $x = 31$ cm.