a canon fires shell at 300m/s. at what angle must it aim to hit a target 5.0km away?
To find the angle at which the cannon must be aimed to hit a target 5.0 km away, we can use the kinematic equations of motion.
We can divide the problem into horizontal and vertical components. The horizontal component of the velocity remains constant, while the vertical component is subject to acceleration due to gravity.
Given:
Initial velocity, u = 300 m/s
Distance to the target, s = 5.0 km = 5000 m
Acceleration due to gravity, g = 9.8 m/s²
Horizontal component of velocity (Vx) = u * cos(theta)
Vertical component of velocity (Vy) = u * sin(theta)
Using the equation of motion for the horizontal component:
s = Vx * t
5000 = (u * cos(theta)) * t
Simplifying, t = 5000 / (u * cos(theta))
Now, using the equation of motion for the vertical component:
y = Vy * t + (0.5) * g * t²
Where y is the vertical displacement (which is zero since the target is at the same height as the cannon).
0 = (u * sin(theta)) * t + (0.5) * g * t²
0 = (u * sin(theta)) * (5000 / (u * cos(theta))) + (0.5) * g * (5000 / (u * cos(theta)))²
Simplifying further, we get:
0 = 5000 * tan(theta) + (0.5) * g * (5000 / u)²
Now we can solve this equation to find the angle (theta) at which the cannon must be aimed.
1. Substitute the given values: u = 300 m/s and g = 9.8 m/s².
2. Rearrange the equation to solve for tan(theta).
3. Use the inverse tangent (arctan) function to find the angle (theta).
Let's calculate the angle using these steps: