A person walking with a speed of 1.44 m/sec releases a ball from a height of h = 1.28 m above the ground. Use the point on the ground, directly below where the ball is initially released at the origin of your coordinate system.
(a) What is the ball's position at 0.25 seconds after it is released?
(b) What is the ball's position at 0.5 seconds after it is released?
(c) What is the ball's total velocity, speed and direction of motion at 0.5 seconds after it is released?
cnat
To solve these questions, we need to use the kinematic equations for motion under constant acceleration. The key equation we'll be using is:
y = y0 + v0t + (1/2)at^2
Where:
y = final position
y0 = initial position
v0 = initial velocity
t = time
a = acceleration
In this case, the ball is dropped, so the initial velocity (v0) is 0 m/s. The acceleration (a) is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2. The initial position (y0) is given as 1.28 m.
(a) To find the ball's position at 0.25 seconds after it is released, we can substitute the values into the equation:
y = 1.28 m + 0 m/s * 0.25 s + (1/2)(9.8 m/s^2)(0.25 s)^2
Simplifying, we get:
y = 1.28 m + 0 + (1/2)(9.8 m/s^2)(0.0625 s^2)
y = 1.28 m + 0 + 0.305 m
y ≈ 1.585 m
So, the ball's position at 0.25 seconds after it is released is approximately 1.585 m above the ground.
(b) To find the ball's position at 0.5 seconds after it is released, we can use the same equation:
y = 1.28 m + 0 m/s * 0.5 s + (1/2)(9.8 m/s^2)(0.5 s)^2
Simplifying, we get:
y = 1.28 m + 0 + (1/2)(9.8 m/s^2)(0.25 s^2)
y = 1.28 m + 0 + 0.305 m
y ≈ 1.585 m
So, the ball's position at 0.5 seconds after it is released is also approximately 1.585 m above the ground.
(c) To determine the ball's total velocity, speed, and direction of motion at 0.5 seconds after it is released, we need to find the velocity of the ball at that time. We can use another kinematic equation:
v = v0 + at
Plugging the values into the equation, we get:
v = 0 m/s + (9.8 m/s^2)(0.5 s)
v = 4.9 m/s
The ball's velocity at 0.5 seconds is 4.9 m/s.
To find the speed, we simply take the absolute value of the velocity:
speed = |v| = |4.9 m/s|
speed = 4.9 m/s
The ball's speed at 0.5 seconds is 4.9 m/s.
Since the velocity is positive, the direction of motion is downward.
So, at 0.5 seconds after it is released, the ball's position is approximately 1.585 m above the ground, its velocity is 4.9 m/s downward, and its speed is 4.9 m/s.