A ball thrown vertically upwards is caught by the thrower 8 seconds afterwards . (take g =10) When the ball is caught by the thrower: what is its displacement from the initial position?
b) what is the initial velocity?
c) To what height does the ball rise?
d) With regard to (c) at the greatest height reached, what is the velocity of the ball?
a. displacement = 0, it is at the same position as when it started.
b. df=di+vi*t-1/2 g t^2
df=di=0 solve for vi. I have no idea why you were told to use g=10, it is not that value anywhere on Earth. Nor is PI equal to 3
c. at the top, velocity is zero, so vf=0=vi-gt hwere t=4sec, and vi equal to what you found in b.
To find the answers to these questions, we can use the equations of motion for objects in freefall. The key equation we'll be using for each question is:
Final velocity (v) = Initial velocity (u) + (acceleration (a) x time (t))
Remember that since the ball is thrown vertically upwards, the acceleration will be negative (-10).
Now let's solve each question step by step:
a) What is the displacement from the initial position when the ball is caught by the thrower?
To find the displacement, we need to calculate the distance traveled by the ball during the entire motion. Since the ball goes up and then comes back down, the total displacement will be zero. This means that when the ball is caught, its displacement from the initial position will also be zero.
b) What is the initial velocity?
To find the initial velocity, we can use the equation:
Final velocity (v) = Initial velocity (u) + (acceleration (a) x time (t))
Since the final velocity is zero when the ball is caught, and the time is 8 seconds, we can substitute these values into the equation:
0 = u + (-10 x 8)
Simplifying the equation, we get:
0 = u - 80
Rearranging the equation to solve for u (initial velocity), we find:
u = 80
Therefore, the initial velocity of the ball is 80 m/s.
c) To what height does the ball rise?
To find the height reached by the ball, we need to calculate the maximum height it reaches during its motion. To do this, we can use the equation:
Final velocity squared (v²) = Initial velocity squared (u²) + (2 x acceleration (a) x displacement (s))
Since the final velocity at the maximum height will be zero, the equation becomes:
0 = u² + (2 x -10 x s)
Simplifying the equation, we get:
0 = u² - 20s
Substituting the value of u (80) into the equation, we get:
0 = 80² - 20s
Simplifying further, we find:
6400 = 20s
Dividing both sides of the equation by 20, we get:
s = 320
Therefore, the ball rises to a height of 320 meters.
d) With regard to (c), at the greatest height reached, what is the velocity of the ball?
At the greatest height reached, the velocity of the ball will be zero, as it momentarily comes to a stop before starting its descent back to the thrower.