A 375 g ball is thrown vertically upward with a speed of 3.0 m/s. How high did it go? How fast is it moving when it is 0.25 m high?
How fast is it moving when it is 0.25 m high?
The speed of the ball when it is 0.25 m high can be calculated using the equation for kinetic energy: KE = 1/2 mv^2.
KE = 1/2 (375 g)(3.0 m/s)^2 = 5.625 J
KE = 1/2 (375 g)(v^2)
v = sqrt(11.25 J/375 g) = 2.5 m/s
Therefore, the speed of the ball when it is 0.25 m high is 2.5 m/s.
To find out how fast the ball is moving when it is 0.25 m high, we can use the principles of kinematics, specifically the equations of motion.
First, we need to determine the initial velocity of the ball. Since the ball is thrown vertically upward, it has an initial velocity of +3.0 m/s (taking upward direction as positive).
Next, we can use the equation of motion that relates initial velocity, final velocity, acceleration, and displacement:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity
vi = initial velocity
a = acceleration
d = displacement
In this case, we are looking for the final velocity when the ball is 0.25 m high, so we can rewrite the equation as:
vf^2 = vi^2 + 2ad
Simplifying further, we get:
vf^2 = (3.0 m/s)^2 + 2(-9.8 m/s^2)(-0.25 m)
Calculating the equation, we find:
vf^2 = 9 m^2/s^2 + 4.9 m^2/s^2
vf^2 = 13.9 m^2/s^2
Taking the square root of both sides, we find:
vf ≈ ±3.73 m/s
Since the ball is moving upward, the final velocity will be negative. Thus, the speed at which the ball is moving when it is 0.25 m high is approximately 3.73 m/s, in the downward direction.
To find the height the ball went and its velocity when it reaches 0.25 m, we'll need to use kinematic equations.
Given:
- Mass of the ball (m) = 375 g = 0.375 kg
- Initial velocity (u) = 3.0 m/s
- Final velocity (v) = ?
- Height (h) = ?
1. Finding the height the ball went:
First, we need to find the time taken (t) to reach the maximum height. We can use the equation v = u + at, where a is the acceleration due to gravity (-9.8 m/s²) in this case.
Initial velocity (u) = 3.0 m/s
Final velocity (v) = 0 m/s (at maximum height)
Acceleration (a) = -9.8 m/s²
v = u + at
0 = 3.0 - 9.8t
Solving for t:
9.8t = 3.0
t = 3.0 / 9.8
t ≈ 0.3061 s
Now, we can find the height (h) using the equation h = ut + (1/2)at²:
h = (3.0)(0.3061) + (0.5)(-9.8)(0.3061)²
h ≈ 0.4615 m
Therefore, the ball went approximately 0.4615 m high.
2. Finding the velocity when the height is 0.25 m:
Now, we need to find the velocity (v) when the ball is at a height of 0.25 m. We can use the same kinematic equation h = ut + (1/2)at²:
h = 0.25
u = 3.0 m/s
a = -9.8 m/s²
t = ?
Rearranging the equation:
0.25 = (3.0)(t) + (0.5)(-9.8)(t)²
This is a quadratic equation in terms of t. Solving it will give us the time at which the height is 0.25 m. The solution will have two values, one for the upward motion and one for the downward motion. We'll only consider the positive value since we're looking for the time when the ball is moving upward.
Solving the equation, we get t ≈ 0.2665 s.
Now, we can find the velocity (v) when the height is 0.25 m by substituting this value of t into the equation v = u + at:
v = 3.0 + (-9.8)(0.2665)
v ≈ -0.645 m/s
Therefore, the ball is moving approximately -0.645 m/s when its height is 0.25 m. Note that the negative sign indicates it is moving downward.