A ball is thrown straight upward with a speed of +12 m/s.
A) what's the ball's acceleration just after it is thrown?
B) how much time does it take for the ball to rise to its maximum height?
C) what's the approximate maximum height the ball reaches?
To answer these questions, we need to understand the principles of motion and apply the relevant equations. Let's break down each question step-by-step:
A) To find the ball's acceleration just after it is thrown, we can use the equation of motion:
v = u + at
where:
v = final velocity (unknown)
u = initial velocity (12 m/s)
a = acceleration (unknown)
t = time (unknown)
Since we are looking for acceleration, and the ball is thrown straight up, we know that the acceleration due to gravity is acting in the opposite direction to the velocity. Therefore, we can assume the acceleration is -9.8 m/s^2 (taking the value of acceleration due to gravity).
Substituting the known values into the equation, we get:
0 (final velocity) = 12 m/s (initial velocity) + (-9.8 m/s^2) (acceleration) * t
Simplifying the equation, we have:
-12 m/s = -9.8 m/s^2 * t
Solving for acceleration:
acceleration = -12 m/s / t
Therefore, the ball's acceleration just after it is thrown is given by -12 m/s divided by the time it takes to reach that speed.
B) To determine the time taken for the ball to rise to its maximum height, we can use the following equation of motion:
v = u + at
where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (12 m/s)
a = acceleration (-9.8 m/s^2)
t = time (unknown)
Substituting the known values, we have:
0 m/s = 12 m/s + (-9.8 m/s^2) * t
Simplifying the equation, we get:
-12 m/s = -9.8 m/s^2 * t
Solving for time:
t = -12 m/s / -9.8 m/s^2
Therefore, the time it takes for the ball to rise to its maximum height is approximately 1.22 seconds.
C) To find the approximate maximum height the ball reaches, we can use the following equation of motion:
s = ut + (1/2)at^2
where:
s = displacement (unknown)
u = initial velocity (12 m/s)
t = time (1.22 seconds, as calculated in part B)
a = acceleration (-9.8 m/s^2)
Substituting the known values, we have:
s = (12 m/s) * (1.22 s) + (1/2) * (-9.8 m/s^2) * (1.22 s)^2
Simplifying the equation, we get:
s = 14.64 m - 8.50 m
Therefore, the approximate maximum height the ball reaches is approximately 6.14 meters.
A) To find the ball's acceleration just after it is thrown, we need to use the equation for acceleration:
acceleration = (final velocity - initial velocity) / time
Since the ball is thrown straight upward, its final velocity at the top of its trajectory will be 0 m/s. The initial velocity is +12 m/s. Let's assume the time it takes for the ball to reach its peak is t.
acceleration = (0 - 12) / t
acceleration = -12 / t
Therefore, the ball's acceleration just after it is thrown is -12/t m/s².
B) To find the time it takes for the ball to rise to its maximum height, we need to find the time it takes for the ball to reach its peak. We know that the final velocity at the top of its trajectory is 0 m/s and the initial velocity is +12 m/s. Using the kinematic equation:
final velocity = initial velocity + (acceleration * time)
We substitute the values we know:
0 = 12 + (-12/t)
Simplifying further:
-12 = -12/t
Cancelling out the negative signs:
12 = 12/t
Solving for t:
t = 1 second
Therefore, it takes 1 second for the ball to rise to its maximum height.
C) To find the approximate maximum height the ball reaches, we can use the equation:
height = (initial velocity * time) + (0.5 * acceleration * time²)
Plugging in the values we know:
height = (12 * 1) + (0.5 * -12/t * 1²)
Simplifying further:
height = 12 - 6
height = 6 meters
Therefore, the approximate maximum height the ball reaches is 6 meters.