A 3.00 kg ball is thrown vertically into the air with an initial velocity of 15.0 m/s. The maximum height of the ball is
The ball will stop rising when the potential energy gain (M g H) equals the initial kinetic energy (M V^2/2).
Solve for H.
This is not college level physics
To find the maximum height of the ball, we can use the equations of motion.
First, we need to find the time it takes for the ball to reach its highest point. We can use the equation:
vf = vi + at
Here, vf is the final velocity (which is 0 m/s at the highest point), vi is the initial velocity (15.0 m/s), a is the acceleration (which is the acceleration due to gravity, approximately -9.8 m/s^2), and t is the time.
So, plugging in the values:
0 = 15.0 + (-9.8)t
Solving for t:
9.8t = 15.0
t ≈ 15.0 / 9.8
t ≈ 1.53 seconds
Now, we can use the equation for displacement to find the maximum height:
y = vi * t + (1/2) * a * t^2
Here, y is the displacement (which is the maximum height), vi is the initial velocity (15.0 m/s), a is the acceleration (which is -9.8 m/s^2), and t is the time (1.53 seconds).
Plugging in the values:
y = 15.0 * 1.53 + (1/2) * (-9.8) * (1.53)^2
y ≈ 23.1 - 11.8
y ≈ 11.3 meters
Therefore, the maximum height of the ball is approximately 11.3 meters.