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

How high the arrow goes if fired upward can be found from energy conservation principle:

$$mgh=\frac{1}{2}mv^2,\\ \space\\ h=\frac{v^2}{2g}=\frac{5^2}{2\cdot9.8}=1.28\text{ m}.$$

Assume the archer was 5 meters above the ground. Therefore, the maximum height the arrow reaches is 6.28 m. Then the arrow falls down from rest and increases its speed. The height of 3 meters relates to 3.28 meters of free falling, therefore, the speed will be

$$u=\sqrt{2gH}=\sqrt{2\cdot9.8\cdot3.28}=8.02\text{ m/s}.$$