A cannon at ground level launches a cannon ball at an angle of 30 degrees with a velocity of 28.3m/s. How far will the cannon ball travel?

To find how far the cannonball will travel, we can break down its initial velocity into horizontal and vertical components.

The horizontal component (Vx) represents the velocity in the x-axis, and the vertical component (Vy) represents the velocity in the y-axis.

Given:
Angle of launch = 30 degrees (θ)
Initial velocity = 28.3 m/s

First, we need to find the horizontal component of the velocity (Vx). We can use the formula:

Vx = V * cos(θ)

where V is the initial velocity and θ is the angle of launch.

Vx = 28.3 m/s * cos(30°) = 24.5 m/s

Next, we need to find the vertical component of the velocity (Vy). We can use the formula:

Vy = V * sin(θ)

Vy = 28.3 m/s * sin(30°) = 14.15 m/s

Now, we can determine the time of flight (t) for the cannonball, which is the time it takes for the vertical velocity component to reach zero. We can use the formula:

t = (2 * Vy) / g

where g is the acceleration due to gravity, approximately 9.8 m/s².

t = (2 * 14.15 m/s) / 9.8 m/s² = 2.88 s

Finally, we can calculate the horizontal distance traveled (d) by multiplying the horizontal velocity (Vx) by the time of flight (t):

d = Vx * t

d = 24.5 m/s * 2.88 s = 70.56 m

Therefore, the cannonball will travel approximately 70.56 meters.

figure horizonatl and initial vertical velocity.

from vertical velocity,

hf=hi+1/2 g t^2+vivertical*t solve for t.

then knowing t, howfar= horizontalvelocity*time