A 1.2 kg cannon ball is shot off a platform (1) at 10 m/s. It hits a platform (2) and begins to roll without bouncing. The coefficient of friction between platform (2) and the ball is .3. How far does the ball land from platform (2)? How long does it take to hit the ground?
You need a figure to show what is going on. It should show the difference in platform heights and how far apart they are. You have already stated that the cannonball "hits" platform 2. How far below is the ground?
To find the distance the ball lands from platform (2), we need to calculate the total horizontal distance traveled by the ball.
First, let's find the time it takes for the ball to hit the ground. Since there is no bouncing and the initial vertical velocity is zero, we can use the equation:
\[ h = \frac{1}{2} g t^2 \]
Where:
- h is the height (initially at platform 1)
- g is the acceleration due to gravity (assuming 9.8 m/s^2)
- t is the time
Rearranging the equation, we get:
\[ t^2 = \frac{2h}{g} \]
Let's assume that the height (h) is the same as the height of platform 1. Therefore, we have:
\[ t^2 = \frac{2 \cdot 0}{9.8} = 0 \]
This means that it takes zero seconds for the ball to hit the ground.
Now, let's find the horizontal distance traveled by the ball. The horizontal distance can be calculated using the equation:
\[ d = v \cdot t \]
Where:
- d is the horizontal distance (what we want to find)
- v is the initial horizontal velocity of the ball (10 m/s)
- t is the time (zero in this case)
Using the equation, we get:
\[ d = 10 \cdot 0 = 0 \]
This means that the ball lands at platform 2 itself (zero distance from platform 2).
In summary, the ball will land at platform 2 itself and it will take zero seconds to hit the ground.