A baseball is hit straight up in the air with an initial velocity of 26.7 m/s. How long in seconds does it stay in the air? (Hint: What is the overall distance of the ball?)
9.8 ballM/s^2
To find the time the baseball stays in the air, we can use the fact that the vertical displacement of the ball when it reaches its peak is equal to zero. We can use the kinematic equation:
Δy = V₀y * t + (1/2) * a * t²
where:
Δy is the vertical displacement (which is 0 at the peak),
V₀y is the initial vertical velocity (26.7 m/s upwards),
a is the acceleration due to gravity (-9.8 m/s²), and
t is the time we're trying to find.
Since the ball reaches the peak when Δy = 0, we can rewrite the equation as:
0 = V₀y * t + (1/2) * a * t²
Simplifying the equation further, we have:
(1/2) * a * t² = -V₀y * t
Dividing both sides by t, we get:
(1/2) * a * t = -V₀y
Now, substituting the given values:
(1/2) * (-9.8 m/s²) * t = -26.7 m/s
Simplifying the equation:
-4.9 t = -26.7
Dividing both sides by -4.9:
t = -26.7 / -4.9
t ≈ 5.449 seconds
Therefore, the baseball stays in the air for approximately 5.449 seconds.