Question

A cannonball is launched at a velocity of 25 m/s in a direction making an angle of 30° upward with the horizontal. What is the direction of its velocity just before it hits the ground?

Answer 1

To determine the direction of the velocity of the cannonball just before it hits the ground, we can break down the initial velocity into its horizontal and vertical components.

The horizontal component of the velocity remains constant throughout the motion and does not change. In this case, the horizontal component can be found using the equation:

$V_{\text{horizontal}}=V\cdot \cos (\theta )$,

where $V$ is the initial velocity of 25 m/s and $\theta$ is the angle of 30°. Substituting the values into the equation:

$V_{\text{horizontal}}=25\cdot \cos (30°)$.

Using the cosine of 30°, which is $\frac{\sqrt{3}}{2}$, we get:

Vhorizontal=25⋅32=2532V_{ ext{horizontal}} = 25 cdot frac{sqrt{3}}{2} = frac{25sqrt{3}}{2}Vhorizontal​=25⋅23​​<span