Question

a 300g ball rolls down a hill that is 1.4 m high. At the bottom of the hill, the ball has a speed of 7.6 m/s. Determine the speed of the ball at the top of the hill

Answer 1

Energy conservation law tells us that

$$\frac{mu^2}{2} +mgh = \frac{mv^2}{2}$$

where v is speed at the bottom of the hill, u is the speed at the top of the hill. Therefore

$$u =\sqrt{v^2-2gh} = \sqrt{7.6^2-2\cdot 9.8 \cdot 1.4} = 5.5 \,\text{m/s}$$