Use these values of initial position and initial velocity in the following questions.
Initial position: ___1.0___ m above ground
Initial velocity: ___6.3 __ m/s, up
Question What is the magnitude and direction of the acceleration as the ball goes up?
What is the magnitude and direction of the acceleration as the ball goes down?
Problem Try to answer the following questions:
(a) What is the maximum height above ground reached by the ball?
(b) What are the magnitude and the direction of the velocity of the ball just
before it hits the ground?
Show Your
Problem
Solving
Steps: Show these below: Choose origin, coordinate direction
Inventory List – What is known?
Write the kinematics equation(s) and solution of Part (a):
Write the kinematics equation(s) and solution of Part (b):
To answer the questions, we need to understand the kinematics equations that relate position, velocity, acceleration, and time. The kinematics equations are as follows:
1. Position equation:
- h = h0 + v0t + (1/2)at^2
- Here, h is the final position, h0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time.
2. Velocity equation:
- v = v0 + at
- Here, v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.
Now let's solve the problems:
(a) What is the maximum height above ground reached by the ball?
To find the maximum height, we need to determine the time at which the velocity becomes zero. At the maximum height, the ball momentarily stops before falling back down.
Using the velocity equation, we can set v = 0 and solve for t:
0 = v0 + at
0 = 6.3 + (-9.8)t
Solving for t, we get t = 6.3/9.8 ≈ 0.6439 seconds.
Now we can substitute this time value into the position equation to find the maximum height:
h = h0 + v0t + (1/2)at^2
h = 1.0 + (6.3)(0.6439) + (1/2)(-9.8)(0.6439)^2
h ≈ 4.04 meters
Therefore, the maximum height reached by the ball is approximately 4.04 meters above the ground.
(b) What are the magnitude and direction of the velocity of the ball just before it hits the ground?
To find the velocity just before the ball hits the ground, we need to determine the time it takes for the ball to reach the ground from its initial position.
Using the position equation, we can set h = 0 (ground level) and solve for t:
0 = h0 + v0t + (1/2)at^2
0 = 1.0 + (6.3)t + (1/2)(-9.8)t^2
Solving this quadratic equation, we get t ≈ 1.3812 seconds (rounded to four decimal places).
Now we can substitute this time value into the velocity equation to find the velocity just before hitting the ground:
v = v0 + at
v = 6.3 + (-9.8)(1.3812)
v ≈ -13.53 m/s (rounded to two decimal places)
The magnitude of the velocity (speed) is approximately 13.53 m/s, and the direction is downward.