Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
Not sure.can you give me a clue?
Answer:
r = 58.44 [m]
Explanation:
To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.
a = v²/r
where:
a = centripetal acceleration = 15.4 [m/s²]
v = tangential speed = 30 [m/s]
r = radius or distance [m]
r = v²/a
r = 30²/15.4
r = 58.44 [m]
Answer/Explanation: Speed and direction can change with time. When you throw a ball into the air, it leaves your hand at a certain speed. As the ball rises, it slows down. Then, as the ball falls back toward the ground, it speeds up again. When the ball hits the ground, its direction of motion changes and it bounces back up into the air.
