A ball is thrown directly upward with an initial velocity of 14 m/s. If the ball is released from an initial height of 2.8 m above ground, how long is the ball in the air before landing on the ground? Ignore air drag.
Vf = Vo^2 + 2gd,
d(up) = (Vf^2 - Vo^2) / 2g,
d(up) = (0 - (14)^2) / -19.6 = 10m.
D = 2.8 + 10 = 12.8m above ground.
Vf = Vo + gt,
t(up) = (Vf - Vo) / g,
t(up) = (0 - 14) / -9.8 = 1.43s.
D = Vo*t + 0.5gt^2 = 12.8,
0 + 0.5*9.8t^2 = 12.8,
4.9t^2 = 12.8,
t^2 = 2.61,
t(down) = 1.62s.
T = 1.43 + 1.62 = 3.05s. in air.
To find the time it takes for the ball to land on the ground, we can use the equation of motion for vertically thrown objects. This equation relates the displacement, initial velocity, time, and acceleration due to gravity.
1. The initial velocity, u, is given as 14 m/s.
2. The acceleration due to gravity, a, is -9.8 m/s^2 (negative because it is acting in the opposite direction to the initial velocity).
3. The displacement, s, is the distance between the initial height and the ground, which is 2.8 m in this case.
4. The time, t, is what we are trying to find.
Using the equation:
s = ut + (1/2)at^2
Substituting the given values into the equation:
2.8 = 14t + (1/2)(-9.8)t^2
Simplifying the equation:
2.8 = 14t - 4.9t^2
Rearranging and combining like terms to form a quadratic equation:
4.9t^2 - 14t + 2.8 = 0
Now, we can solve this quadratic equation to find the values of t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 4.9, b = -14, and c = 2.8. Plugging them into the formula:
t = (-(-14) ± √((-14)^2 - 4 * 4.9 * 2.8)) / (2 * 4.9)
Simplifying the equation further:
t = (14 ± √(196 - 54.88)) / 9.8
t = (14 ± √141.12) / 9.8
Now, we can calculate the two possible values for t using the positive and negative square root:
t₁ = (14 + √141.12) / 9.8
t₂ = (14 - √141.12) / 9.8
Calculating the values:
t₁ ≈ 2.46 seconds
t₂ ≈ -0.37 seconds (discarded as it is negative)
The ball will be in the air for approximately 2.46 seconds before landing on the ground.