A cannon ball is fired horizontally with an initial speed of 75 m/s from the top of a 50 m tall building. When the ball hits the ground, its velocity vector makes an angle with the horizontal. What is the angle?
See previous post.
To find the angle at which the velocity vector makes with the horizontal when the cannonball hits the ground, we need to consider the horizontal and vertical components of the motion.
First, let's look at the vertical motion. The cannonball is fired horizontally, so there is no initial vertical velocity. The only force acting on the ball in the vertical direction is gravity, which causes it to accelerate downwards. The distance covered vertically is the height of the building, which is 50 m.
Using the equation of motion for vertical motion:
Final velocity squared = Initial velocity squared + 2 * acceleration * distance
Final velocity = 0 (when the ball hits the ground)
Initial velocity = 0 (since there is no initial vertical velocity)
Simplifying the equation, we have:
0 = 0 + 2 * acceleration * 50
Solving for the acceleration due to gravity, we find:
acceleration = 0 / (2 * 50) = 0 m/s²
Since the acceleration is 0 m/s², this means that the ball is in freefall vertically (ignoring air resistance).
Now, let's consider the horizontal motion. The ball is fired horizontally with an initial speed of 75 m/s. There is no acceleration in the horizontal direction, so the ball continues to move horizontally at a constant speed of 75 m/s until it hits the ground.
Since the velocity is constant in the horizontal direction, the angle that the velocity vector makes with the horizontal is 0 degrees.
Therefore, the angle the velocity vector makes with the horizontal when the ball hits the ground is 0 degrees.