A softball player tosses a ball into the air with an initial velocity of 11.0 m/s, as shown in the diagram. What will be the ball’s maximum height?
v^2 = 2gh
To find the maximum height of the ball, we can use the equations of motion. Assuming there is no air resistance, we can use the following equation to determine the maximum height:
v^2 = u^2 + 2as
Where:
- v is the final velocity (which will be zero at the maximum height)
- u is the initial velocity (11.0 m/s in this case)
- a is the acceleration (which is -9.8 m/s^2, considering the acceleration due to gravity)
- s is the displacement (which we want to find, the maximum height)
Since the final velocity is zero at the maximum height, we can rewrite the equation as:
0 = (11.0 m/s)^2 + 2(-9.8 m/s^2)s
Simplifying the equation:
0 = 121 m^2/s^2 - 19.6 m/s^2 s
Rearranging the equation to isolate s:
19.6 m/s^2 s = 121 m^2/s^2
Dividing both sides of the equation by 19.6 m/s^2:
s = 121 m^2/s^2 / 19.6 m/s^2
s ≈ 6.17 m
Therefore, the ball will reach a maximum height of approximately 6.17 meters.
To find the ball's maximum height, we need to understand the projectile motion of the softball. In projectile motion, the vertical motion is influenced by gravity, while the horizontal motion remains constant.
To determine the maximum height, we can use the equation for vertical displacement during projectile motion:
Δy = (v₀y²) / (2g)
Where:
Δy is the vertical displacement (maximum height)
v₀y is the initial vertical velocity
g is the acceleration due to gravity (9.8 m/s²)
In this case, the initial vertical velocity (v₀y) is given as 11.0 m/s. Thus, we can substitute the values into the equation:
Δy = (11.0 m/s)² / (2 * 9.8 m/s²)
Simplifying the equation:
Δy = 121.0 m²/s² / 19.6 m/s²
Δy = 6.17 m
Therefore, the maximum height of the softball will be approximately 6.17 meters.