Question

I A cannon tilted upward at${30}^{°}$ fires a cannonball with a speed of $100\text{m}/\text{s}$. What is the component of the cannonball's velocity parallel to the ground?

Short answer

Component of the cannon ball's velocity parallel to the ground $={\text{v}}_{\text{x}}$

$=86.6\text{m}/\text{s}$

Step-by-step solution

Step.1

Step.2.

Angle made by the cannon ball with the ground$\theta ={30}^{°}$

Magnitude of velocity$\text{v}=100\text{m}/\text{s}$

Velocity $\overrightarrow{\text{v}}={\text{v}}_{\text{x}}\widehat{i}+{\text{v}}_{\text{y}}\widehat{j}$

$\begin{array}{l}=(100\text{m}/\text{s})\cos {30}^{°}\widehat{i}+(100\text{m}/\text{s})\sin {30}^{°}\widehat{j}\\ =(86.6\text{m}/\text{s})\widehat{i}+(50\text{m}/\text{s})\widehat{j}\end{array}$

Component of the cannon ball's velocity parallel to the ground $={\text{v}}_{\text{x}}$

$=86.6\text{m}/\text{s}$