Question

An arrow is fired by an archer at a speed of 5m/s. Its path has the equation of 220st t=−meters after t seconds.
How high does the arrow go and find the speed covered by the rocket when its 3m above the ground?

Answer 1

From the conservation of energy:

$$mgH=0.5mv^2$$

Height of an arrow:

$$H=\frac{v^2}{2g}=\frac{5^2}{2(9.8)}=1.3\ m$$

The speed covered by the rocket when its 3m above the ground:

$$V(3)=\sqrt{V^2-2gh}$$

$$V(3)=\sqrt{220^2-2(9.8)(3)}=219.87\frac{m}{s}$$